\nx3<\/sup> + 9x = \\(\\frac{1}{2}\\) (18x + 54)
\nAnswer:
\nNow that we know about square roots and cube roots, we can combine that knowledge with our knowledge of the properties of equality to begin solving nonlinear equations like x3<\/sup> + 9x = \\(\\frac{1}{2}\\) (18x + 54). Transform the equation until you can determine the positive value of x that makes the equation true.
\nChallenge students to solve the equation independently or in pairs. Have students share their strategy for solving the equation. Ask them to explain each step.
\nx3<\/sup> + 9x = \\(\\frac{1}{2}\\) (18x + 54)
\nx3<\/sup> + 9x = 9x + 27
\nx3<\/sup> + 9x – 9x = 9x – 9x + 27
\nx3<\/sup> = 27
\n\\(\\sqrt[3]{x^{3}}\\) = \\(\\sqrt [ 3 ]{ 27 }\\)
\nx = \\(\\sqrt[3]{3^{3}}\\)
\nx = 3
\nNow, we verify our solution is correct.
\n33<\/sup> + 9(3) = \\(\\frac{1}{2}\\) (18(3) + 54)
\n27 + 27 = \\(\\frac{1}{2}\\) (54 + 54)
\n54 = \\(\\frac{1}{2}\\) (108)
\n54 = 54
\nSince the left side is the same as the right side, our solution is correct.<\/p>\n
Example 2. Let x = – 8. Find the positive value of x that makes each equation true, and then verify your solution is correct. Check: ( – 9)2<\/sup> – 14 = 5( – 9) + 67 – 5( – 9) b. Explain how you solved the equation. Exercise 2. Check: – 11( – 11 – 1) = 121 – ( – 11) Exercise 3. Check: Exercise 4. Check: Exercise 5. Check: – 9( – 9 + 4) – 3 = 4( – 9 + 19.5) Exercise 6. Check: Exercise 7. b. Determine the value of x, and check your answer. Check: Find the positive value of x that makes each equation true, and then verify your solution is correct. Check: Question 2. Check: Question 3. Check: Question 4. Check: Question 5. Check: Question 6. Check: Question 7. Check: Question 8. Let A represent the area of the triangle. Question 9. Check: Question 10. Check: Question 1. Check: Question 2. Check: Engage NY Eureka Math 8th Grade Module 7 Lesson 5 Answer Key Eureka Math Grade 8 Module 7 Lesson 5 Example Answer Key Example 1. x3 + 9x = (18x + 54) Answer: Now that we know about square roots and cube roots, we can combine that knowledge with our knowledge of the properties of … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/82313"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=82313"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/82313\/revisions"}],"predecessor-version":[{"id":137303,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/82313\/revisions\/137303"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=82313"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=82313"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=82313"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nx(x – 3) – 51 = – 3x + 13
\nAnswer:
\nLet\u2019s look at another nonlinear equation. Find the positive value of x that makes the equation true: x(x – 3) – 51 = – 3x + 13.
\nProvide students with time to solve the equation independently or in pairs. Have students share their strategy for solving the equation. Ask them to explain each step.
\nSample response:
\nx(x – 3) – 51 = – 3x + 13
\nx2<\/sup> – 3x – 51 = – 3x + 13
\nx2<\/sup> – 3x + 3x – 51 = – 3x + 3x + 13
\nx2<\/sup> – 51 = 13
\nx2<\/sup> – 51 + 51 = 13 + 51
\nx2<\/sup> = 64
\n\\(\\sqrt{x^{2}}\\) = \u00b1\\(\\sqrt{64}\\)
\nx = \u00b1\\(\\sqrt{64}\\)
\nx = \u00b18
\nNow we verify our solution is correct.
\nProvide students time to check their work.
\nLet x = 8.
\n8(8 – 3) – 51 = – 3(8) + 13
\n8(5) – 51 = – 24 + 13
\n40 – 51 = – 11
\n– 11 = – 11<\/p>\n
\n– 8( – 8 – 3) – 51 = – 3( – 8) + 13
\n– 8( – 11) – 51 = 24 + 13
\n88 – 51 = 37
\n37 = 37
\nNow it is clear that the left side is exactly the same as the right side, and our solution is correct.<\/p>\nEureka Math Grade 8 Module 7 Lesson 5 Exercise Answer Key<\/h3>\n
\nExercise 1.
\na. Solve x2<\/sup> – 14 = 5x + 67 – 5x.
\nAnswer:
\nx2<\/sup> – 14 = 5x + 67 – 5x
\nx2<\/sup> – 14 = 67
\nx2<\/sup> – 14 + 14 = 67 + 14
\nx2<\/sup> = 81
\n\\(\\sqrt{x^{2}}\\) = \u00b1\\(\\sqrt{81}\\)
\nx = \u00b1\\(\\sqrt{81}\\)
\nx = \u00b19<\/p>\n
\n92<\/sup> – 14 = 5(9) + 67 – 5(9)
\n81 – 14 = 45 + 67 – 45
\n67 = 67<\/p>\n
\n81 – 14 = – 45 + 67 + 45
\n67 = 67<\/p>\n
\nAnswer:
\nTo solve the equation, I had to first use the properties of equality to transform the equation into the form of x2<\/sup> = 81. Then, I had to take the square root of both sides of the equation to determine that x = 9 since the number x is being squared.<\/p>\n
\nSolve and simplify: x(x – 1) = 121 – x.
\nAnswer:
\nx(x – 1) = 121 – x
\nx2<\/sup> – x = 121 – x
\nx2<\/sup> – x + x = 121 – x + x
\nx2<\/sup> = 121
\n\\(\\sqrt{x^{2}}\\) = \u00b1\\(\\sqrt{121}\\)
\nx = \u00b1\\(\\sqrt{121}\\)
\nx = \u00b111<\/p>\n
\n11(11 – 1) = 121 – 11
\n11(10) = 110
\n110 = 110<\/p>\n
\n– 11( – 12) = 121 + 11
\n132 = 132<\/p>\n
\nA square has a side length of 3x inches and an area of 324 in2<\/sup>. What is the value of x?
\nAnswer:
\n(3x)2<\/sup> = 324
\n32<\/sup> x2<\/sup> = 324
\n9x2<\/sup> = 324
\n\\(\\frac{9 x^{2}}{9}\\) = \\(\\frac{324}{9}\\)
\nx2<\/sup> = 36
\n\\(\\sqrt{x^{2}}\\) = \\(\\sqrt{36}\\)
\nx = 6<\/p>\n
\n(3(6)) 2<\/sup> = 324
\n182<\/sup> = 324
\n324 = 324
\nA negative number would not make sense as a length, so x = 6.<\/p>\n
\n– 3x3<\/sup> + 14 = – 67
\nAnswer:
\n– 3x3<\/sup> + 14 = – 67
\n– 3x3<\/sup> + 14 – 14 = – 67 – 14
\n– 3x3<\/sup> = – 81
\n\\(\\frac{ – 3 x^{3}}{3}\\) = \\(\\frac{ – 81}{ – 3}\\)
\nx3<\/sup> = 27
\n\\(\\sqrt[3]{x^{3}}\\) = \\(\\sqrt [ 3 ]{ 27 }\\)
\nx = 3<\/p>\n
\n– 3(3)3<\/sup> + 14 = – 67
\n– 3(27) + 14 = – 67
\n– 81 + 14 = – 67
\n– 67 = – 67<\/p>\n
\nx(x + 4) – 3 = 4(x + 19.5)
\nAnswer:
\nx(x + 4) – 3 = 4(x + 19.5)
\nx2<\/sup> + 4x – 3 = 4x + 78
\nx2<\/sup> + 4x – 4x – 3 = 4x – 4x + 78
\nx2<\/sup> – 3 = 78
\nx2<\/sup> – 3 + 3 = 78 + 3
\nx2<\/sup> = 81
\n\\(\\sqrt{x^{2}}\\) = \u00b1\\(\\sqrt{81}\\)
\nx = \u00b19<\/p>\n
\n9(9 + 4) – 3 = 4(9 + 19.5)
\n9(13) – 3 = 4(28.5)
\n117 – 3 = 114
\n114 = 114<\/p>\n
\n– 9( – 5) – 3 = 4(10.5)
\n45 – 3 = 42
\n42 = 42<\/p>\n
\n216 + x = x(x2<\/sup> – 5) + 6x
\nAnswer:
\n216 + x = x(x2<\/sup> – 5) + 6x
\n216 + x = x3<\/sup> – 5x + 6x
\n216 + x = x3<\/sup> + x
\n216 + x – x = x3<\/sup> + x – x
\n216 = x3<\/sup>
\n\\(\\sqrt [ 3 ]{ 216 }\\) = \\(\\sqrt[3]{x^{3}}\\)
\n6 = x<\/p>\n
\n216 + 6 = 6(62<\/sup> – 5) + 6(6)
\n222 = 6(31) + 36
\n222 = 186 + 36
\n222 = 222<\/p>\n
\na. What are we trying to determine in the diagram below?
\n![\"Engage](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Engage-NY-Math-Grade-8-Module-7-Lesson-5-Exercise-Answer-Key-1.png\")
\nAnswer:
\nWe need to determine the value of x so that its square root, multiplied by 4, satisfies the equation
\n52<\/sup> + (4\\(\\sqrt{x}\\))2<\/sup> = 112<\/sup>.<\/p>\n
\nAnswer:
\n52<\/sup> + (4\\(\\sqrt{x}\\))2<\/sup> = 112<\/sup>
\n25 + 42<\/sup> (\\(\\sqrt{x}\\))2<\/sup> = 121
\n25 – 25 + 42<\/sup> (\\(\\sqrt{x}\\))2<\/sup> = 121 – 25
\n16x = 96
\n\\(\\frac{16x}{16}\\) = \\(\\frac{96}{16}\\)
\nx = 6
\nThe value of x is 6.<\/p>\n
\n52<\/sup> + (4\\(\\sqrt{6}\\))2<\/sup> = 112<\/sup>
\n25 + 16(6) = 121
\n25 + 96 = 121
\n121 = 121<\/p>\nEureka Math Grade 8 Module 7 Lesson 5 Problem Set Answer Key<\/h3>\n
\nQuestion 1.
\nx2<\/sup> (x + 7) = \\(\\frac{1}{2}\\) (14x2<\/sup> + 16)
\nAnswer:
\nx2<\/sup> (x + 7) = \\(\\frac{1}{2}\\) (14x2<\/sup> + 16)
\nx3<\/sup> + 7x2<\/sup> = 7x2<\/sup> + 8
\nx3<\/sup> + 7x2<\/sup> – 7x2<\/sup> = 7x2<\/sup> – 7x2<\/sup> + 8
\nx3<\/sup> = 8
\n\\(\\sqrt[3]{x^{3}}\\) = \\(\\sqrt [ 3 ]{ 8 }\\)
\nx = 2<\/p>\n
\n22<\/sup> (2 + 7) = \\(\\frac{1}{2}\\) (14(22<\/sup> ) + 16)
\n4(9) = \\(\\frac{1}{2}\\) (56 + 16)
\n36 = \\(\\frac{1}{2}\\) (72)
\n36 = 36<\/p>\n
\nx3<\/sup> = 1331 – 1<\/sup>
\nAnswer:
\nx3<\/sup> = 1331 – 1<\/sup>
\n\\(\\sqrt[3]{x^{3}}\\) = \\(\\sqrt[3]{1331^{ – 1}}\\)
\nx = \\(\\sqrt[3]{\\frac{1}{1331}}\\)
\nx = \\(\\sqrt[3]{\\frac{1}{11^{3}}}\\)
\nx = \\(\\frac{1}{11}\\)<\/p>\n
\n(\\(\\left(\\frac{1}{11}\\right)^{3}\\))3<\/sup> = 1331 – 1<\/sup>
\n\\(\\frac{1}{11^{3}}\\) = 1331 – 1<\/sup>
\n\\(\\frac{1}{1331}\\) = 1331 – 1<\/sup>
\n1331 – 1<\/sup> = 1331 – 1<\/sup><\/p>\n
\nDetermine the positive value of x that makes the equation true, and then explain how you solved the equation.
\n\\(\\frac{x^{9}}{x^{7}}\\) – 49 = 0
\nAnswer:
\n\\(\\frac{x^{9}}{x^{7}}\\) – 49 = 0
\nx2<\/sup> – 49 = 0
\nx2<\/sup> – 49 + 49 = 0 + 49
\nx2<\/sup> = 49
\n\\(\\sqrt{x^{2}}\\) = \\(\\sqrt{49}\\)
\nx = 7<\/p>\n
\n72<\/sup> – 49 = 0
\n49 – 49 = 0
\n0 = 0
\nTo solve the equation, I first had to simplify the expression \\(\\frac{x^{9}}{x^{7}}\\) to x2<\/sup>. Next, I used the properties of equality to transform the equation into x2<\/sup> = 49. Finally, I had to take the square root of both sides of the equation to solve for x.<\/p>\n
\nDetermine the positive value of x that makes the equation true.
\n(8x)2<\/sup> = 1
\nAnswer:
\n(8x)2<\/sup> = 1
\n64x2<\/sup> = 1
\n\\(\\sqrt{64^{2}}\\) = \\(\\sqrt{1}\\)
\n8x = 1
\n\\(\\frac{8x}{8}\\) = \\(\\frac{1}{8}\\)
\nx = \\(\\frac{1}{8}\\)<\/p>\n
\n(8(\\(\\frac{1}{8}\\)))2<\/sup> = 1
\n12<\/sup> = 1
\n1 = 1<\/p>\n
\n(9\\(\\sqrt{x}\\))2<\/sup> – 43x = 76
\nAnswer:
\n(9\\(\\sqrt{x}\\))2<\/sup> – 43x = 76
\n92<\/sup> (\u221ax)2<\/sup> – 43x = 76
\n81x – 43x = 76
\n38x = 76
\n\\(\\frac{38x}{38}\\) = \\(\\frac{76}{38}\\)
\nx = 2<\/p>\n
\n(9(\\(\\sqrt{2}\\)))2<\/sup> – 43(2) = 76
\n92<\/sup> (\\(\\sqrt{2}\\))2<\/sup> – 86 = 76
\n81(2) – 86 = 76
\n162 – 86 = 76
\n76 = 76<\/p>\n
\nDetermine the length of the hypotenuse of the right triangle below.
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-8th-Grade-Module-7-Lesson-5-Problem-Set-Answer-Key-1.png\")
\nAnswer:
\n32<\/sup> + 72<\/sup> = x2<\/sup>
\n9 + 49 = x2<\/sup>
\n58 = x2<\/sup>
\n\\(\\sqrt{58}\\) = \\(\\sqrt{x^{2}}\\)
\n\\(\\sqrt{58}\\) = x<\/p>\n
\n32<\/sup> + 72<\/sup> = (\\(\\sqrt{52}\\))2<\/sup>
\n9 + 49 = 58
\n58 = 58
\nSince x = \\(\\sqrt{58}\\), the length of the hypotenuse is \\(\\sqrt{58}\\) mm.<\/p>\n
\nDetermine the length of the legs in the right triangle below.
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-8th-Grade-Module-7-Lesson-5-Problem-Set-Answer-Key-2.png\")
\nAnswer:
\nx2<\/sup> + x2<\/sup> = (14\\(\\sqrt{2}\\))2<\/sup>
\n2x2<\/sup> = 142<\/sup> (\\(\\sqrt{2}\\))2<\/sup>
\n2x2<\/sup> = 196(2)
\n\\(\\frac{2 x^{2}}{2}\\) = \\(\\frac{196(2)}{2}\\)
\nx2<\/sup> = 196
\n\\(\\sqrt{x^{2}}\\) = \\(\\sqrt{196}\\)
\nx = \\(\\sqrt{14^{2}}\\)
\nx = 14<\/p>\n
\n142<\/sup> + 142<\/sup> = (14\\(\\sqrt{2}\\))2<\/sup>
\n196 + 196 = 142<\/sup> (\\(\\sqrt{2}\\))2<\/sup>
\n392 = 196(2)
\n392 = 392
\nSince x = 14, the length of each of the legs of the right triangle is 14 cm.<\/p>\n
\nAn equilateral triangle has side lengths of 6 cm. What is the height of the triangle? What is the area of the triangle?
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-8th-Grade-Module-7-Lesson-5-Problem-Set-Answer-Key-3.png\")
\nAnswer:
\nNote: This problem has two solutions, one with a simplified root and one without. Choose the appropriate solution for your classes based on how much simplifying you have taught them.
\nLet h cm represent the height of the triangle.
\n32<\/sup> + h2<\/sup> = 62<\/sup>
\n9 + h2<\/sup> = 36
\n9 – 9 + h2<\/sup> = 36 – 9
\nh2<\/sup> = 27
\n\\(\\sqrt{h^{2}}\\) = \\(\\sqrt{27}\\)
\nh = \\(\\sqrt{27}\\)
\nh = \\(\\sqrt{3^{3}}\\)
\nh = \\(\\sqrt{3^{2}}\\)\u00d7\\(\\sqrt{3}\\)
\nh = 3\\(\\sqrt{3}\\)<\/p>\n
\nA = \\(\\frac{6(3 \\sqrt{3})}{2}\\))
\nA = 3(3\\(\\sqrt{3}\\))
\nA = 9\\(\\sqrt{3}\\)
\nSimplified: The height of the triangle is 3\\(\\sqrt{3}\\) cm, and the area is 9\\(\\sqrt{3}\\) cm2<\/sup>.
\nUnsimplified: The height of the triangle is \\(\\sqrt{27}\\) cm, and the area is 3\\(\\sqrt{27}\\) cm2<\/sup><\/p>\n
\nChallenge: Find the positive value of x that makes the equation true.
\n(\\(\\frac{1}{2}\\) x)2<\/sup> – 3x = 7x + 8 – 10x
\nAnswer:
\n(\\(\\frac{1}{2}\\) x)2<\/sup> – 3x = 7x + 8 – 10x
\n\\(\\frac{1}{4}\\) x2<\/sup> – 3x = – 3x + 8
\n\\(\\frac{1}{4}\\) x2<\/sup> – 3x + 3x = – 3x + 3x + 8
\n\\(\\frac{1}{4}\\) x2<\/sup> = 8
\n4(\\(\\frac{1}{4}\\)) x2<\/sup> = 8(4)
\nx2<\/sup> = 32
\n\\(\\sqrt{x^{2}}\\) = \\(\\sqrt{32}\\)
\nx = \\(\\sqrt{2^{5}}\\)
\nx = \\(\\sqrt{2^{2}}\\) \u22c5 \\(\\sqrt{2^{2}}\\) \u22c5 \\(\\sqrt{2}\\)
\nx = 4\\(\\sqrt{2}\\)<\/p>\n
\n(\\(\\frac{1}{2}\\) (4\\(\\sqrt{2}\\)))2<\/sup> – 3(4\\(\\sqrt{2}\\)) = 7(4\\(\\sqrt{2}\\)) + 8 – 10(4\\(\\sqrt{2}\\))
\n\\(\\frac{1}{4}\\) (16)(2) – 3(4\\(\\sqrt{2}\\)) = 7(4\\(\\sqrt{2}\\)) – 10(4\\(\\sqrt{2}\\)) + 8
\n\\(\\frac{32}{4}\\) – 3(4\\(\\sqrt{2}\\)) = 7(4\\(\\sqrt{2}\\)) – 10(4\\(\\sqrt{2}\\)) + 8
\n8 – 3(4\\(\\sqrt{2}\\)) = (7 – 10)(4\\(\\sqrt{2}\\)) + 8
\n8 – 3(4\\(\\sqrt{2}\\)) = – 3(4\\(\\sqrt{2}\\)) + 8
\n8 – 8 – 3(4\\(\\sqrt{2}\\)) = – 3(4\\(\\sqrt{2}\\)) + 8 – 8
\n– 3(4\\(\\sqrt{2}\\)) = – 3(4\\(\\sqrt{2}\\))<\/p>\n
\nChallenge: Find the positive value of x that makes the equation true.
\n11x + x(x – 4) = 7(x + 9)
\nAnswer:
\n11x + x(x – 4) = 7(x + 9)
\n11x + x2<\/sup> – 4x = 7x + 63
\n7x + x2<\/sup> = 7x + 63
\n7x – 7x + x2<\/sup> = 7x – 7x + 63
\nx2<\/sup> = 63
\n\\(\\sqrt{x^{2}}\\)) = \\(\\sqrt{63}\\)
\nx = \\(\\sqrt{\\left(3^{2}\\right)(7)}\\)
\nx = \\(\\sqrt{3^{2}}\\) \u22c5 \\(\\sqrt{7}\\)
\nx = 3\\(\\sqrt{7}\\)<\/p>\n
\n11(3\\(\\sqrt{7}\\)) + 3\\(\\sqrt{7}\\) (3\\(\\sqrt{7}\\) – 4) = 7(3\\(\\sqrt{7}\\) + 9)
\n33\\(\\sqrt{7}\\) + 32<\/sup> (\\(\\sqrt{7}\\))2<\/sup> – 4(3\\(\\sqrt{7}\\)) = 21\\(\\sqrt{7}\\) + 63
\n33\\(\\sqrt{7}\\) – 4(3\\(\\sqrt{7}\\)) + 9(7) = 21\\(\\sqrt{7}\\) + 63
\n33\\(\\sqrt{7}\\) – 12\\(\\sqrt{7}\\) + 63 = 21\\(\\sqrt{7}\\) + 63
\n(33 – 12) \\(\\sqrt{7}\\) + 63 = 21\\(\\sqrt{7}\\) + 63
\n21\\(\\sqrt{7}\\) + 63 = 21\\(\\sqrt{7}\\) + 63
\n21\\(\\sqrt{7}\\) + 63 – 63 = 21\\(\\sqrt{7}\\) + 63 – 63
\n21\\(\\sqrt{7}\\) = 21\\(\\sqrt{7}\\)<\/p>\nEureka Math Grade 8 Module 7 Lesson 5 Exit Ticket Answer Key<\/h3>\n
\nFind the positive value of x that makes the equation true, and then verify your solution is correct.
\nx2<\/sup> + 4x = 4(x + 16)
\nAnswer:
\nx2<\/sup> + 4x = 4(x + 16)
\nx2<\/sup> + 4x = 4x + 64
\nx2<\/sup> + 4x – 4x = 4x – 4x + 64
\nx2<\/sup> = 64
\n\\(\\sqrt{x^{2}}\\) = \\(\\sqrt{64}\\)
\nx = 8<\/p>\n
\n82<\/sup> + 4(8) = 4(8 + 16)
\n64 + 32 = 4(24)
\n96 = 96<\/p>\n
\nFind the positive value of x that makes the equation true, and then verify your solution is correct.
\n(4x)3<\/sup> = 1728
\nAnswer:
\n(4x)3<\/sup> = 1728
\n64x3<\/sup> = 1728
\n\\(\\frac{1}{64}\\)(64x3<\/sup>) = (1728)\\(\\frac{1}{64}\\)
\nx3<\/sup> = 27
\n\\(\\sqrt[3]{x^{3}}\\) = \\(\\sqrt [ 3 ]{ 27 }\\)
\nx = 3<\/p>\n
\n(4(3))3<\/sup> = 1728
\n123<\/sup> = 1728
\n1728 = 1728<\/p>\n","protected":false},"excerpt":{"rendered":"