Question

The value of
(4a²-9b²)³+(9b²-16c²)³+(16c²-4a²)³/(2a-3b)²+(3b-4c)³+(4c-2a)³ is
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if he give full solution +20 points​

Answer 1

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Answer 2

Hint.

Let $A=2a,B=3b,C=4c$. We observe that $A^2=4a^2,B^2=9b^2,C^2=16c^2$.

Also, we can observe that the sum of the bases of three cubes equal to zero. We have a suitable identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2+ab+bc+ca)$ since it is a sum of three cubes. If the sum of bases are equal to zero, one of the factors is zero,  $a+b+c=0$. Consequently $a^3+b^3+c^3=3abc$.

Solution.

Given, $\dfrac{(4a^2-9b^2)^3+(9b^2-16c^2)^3+(16c^2-4a^2)^3}{(2a-3b)^3+(3b-4c)^3+(4c-2a)^3}$

$=\dfrac{(A^2-B^2)+(B^2-C^2)^3+(C^2-A^2)^3}{(A-B)^3+(B-C)^3+(C-A)^3}$

$=\dfrac{3(A^2-B^2)(B^2-C^2)(C^2-A^2)}{3(A-B)(B-C)(C-A)}$

$=(A+B)(B+C)(C+A)$

Substituting $A=2a,B=3b,C=4c$ back into the equation,

$=(2a+3b)(3b+4c)(4c+2a)$

$=\boxed{2(2a+3b)(3b+4c)(2c+a)}$

This is the required answer.