Solve: (3 x+4 y)^4-x^4 - Vidyadip

Question

Solve:
$(3 x+4 y)^4-x^4$

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Answer

$(3 x+4 y)^4-x^4$
$=\left[(3 x+4 y)^2\right]^2-\left(x^2\right)^2$
$\left.\left.=\left[(3 x+4 y)^2+x^2\right][3 x+] 4 y\right)^2-x^2\right]$
$\left.=\left[(3 x+4 y)^2+x^2\right][3 x+4 y)+x\right][(3 x+4 y)-x]$
$=\left\{(3 x+4 y)^2+x^2\right\}(3 x+4 y+x)(3 x+y-x)$
$=\left\{(3 x+4 y)^2+x^2\right\}(4 x+4 y)(2 x+4 y)$
$\left.=\{3 x+4 y)^2+x^2\right\} 4(x+y) 2(x+2 y)$
$=8\left\{(3 x+4 y)^2+x^2\right\}(x+y)(x+2 y)$

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