Question

the number in the ratio 2:3:4 the sum of their cubes is 33957 the number are​

Answer 1

$\tt\purple{\underbrace{Answer\implies 14,\:21,\:28}}$

$\sf\large\underline{Let:}$

$\tt{\implies The\: numbers\:be\:x}$

$\sf\large\underline{Given:}$

$\tt{\implies The\: ratio\:of\: numbers=2:3:4}$

$\tt{\implies The\:sum\:of\: their\:cubes=33957}$

$\sf\large\underline{To\: Find:}$

$\tt{\implies The\: numbers=?}$

$\sf\large\underline{Solution:}$

• At first we have to find the value of x then calculate the the numbers here. As given in the question, we have to assume the numbers are 2x , 3x, 4x]

$\rm{\implies The\:sum\:of\: their\: cubes=33957}$

$\rm{\implies (2x)^3+(3x)^3+(4x)^3=33957}$

$\rm{\implies 8x^3+27x^3+64x^3=33957}$

$\rm{\implies 99x^3=33957}$

$\rm{\implies x^3=343}$

• Here we have to factorie 343 by prime factorisation method. By factoring we get 7×7×7=7³]

$\rm{\implies x^3=7^3\implies x=7}$

$\sf\large{Hence,}$

$\rm{\implies The\: numbers=2x=2*7=14}$

$\rm{\implies The\: numbers=3x=3*7=21}$

$\rm{\implies The\: numbers=4x=4*7=28}$

$\sf\large\underline{Verification:}$

$\rm{\implies 14^3+21^3+28^3=33957}$

$\rm{\implies 2744+9261+21952=33957}$

$\rm{\implies 33957=33957}$

$\sf\large\underline{Hence,\: Verified:}$

Answer 2

$\sf\red{\underbrace{ Numbers \implies 14,\:21\:and\:28}}$

$\sf{\underline{\underline{Let:}}}$

$\sf{ \: \: \: \: \: three\:numbers\:be\:2x,\:3x\:and\:4x}$

$\sf{\underline{\underline{According\:To\:QuEStiON:}}}$

$\sf{ {(2x)}^{3} + {(3x)}^{3} + {(4x)}^{3} = 33957}$

$\sf{\implies 8{x}^{3} + 27{x}^{3} + 64{x}^{3} = 33957}$

$\sf{\implies 99{x}^{3} = 33957}$

$\sf{\implies {x}^{3} = \dfrac{33957}{99}}$

$\sf{\implies {x}^{3} = 343}$

$\sf{\implies {x}^{3} = 7 \times 7 \times 7 = {(7)}^{3}}$

$\sf{\implies x = 7}$

$\sf{\underline{\underline{Therefore:}}}$

$\sf{The\:greater \:number = 4x = 4 \times 7 = 28}$

$\sf{ And\: other\:numbers:}$

$\sf{3x = 3 \times 7 = 21}$

$\sf{ and\: 2x = 2 \times 7 = 14}$