Question
Solve:
$ (3 x+4 y)^4-x^4$
$ (3 x+4 y)^4-x^4$
✓
Answer
$ (3 x+4 y)^4-x^4$
$= [(3 x+4 y)^2 ]^2- (x^2 )^2$
$= [(3 x+4 y)^2+x^2 ][3 x+] 4 y )^2-x^2 ]$
$= [(3 x+4 y)^2+x^2 ][3 x+4 y)+x ][(3 x+4 y)-x]$
$= \{(3 x+4 y)^2 +x^2\}(3 x+4 y+x)(3 x+y-x)$
$= \{(3 x+4 y)^2+x^2\}(4 x+4 y)(2 x+4 y)$
$=\{3 x+4 y)^2+x^2\} 4(x+y) 2(x+2 y)$
$=8\left\{(3 x+4 y)^2+x^2\right\}(x+y)(x+2 y)$
$= [(3 x+4 y)^2 ]^2- (x^2 )^2$
$= [(3 x+4 y)^2+x^2 ][3 x+] 4 y )^2-x^2 ]$
$= [(3 x+4 y)^2+x^2 ][3 x+4 y)+x ][(3 x+4 y)-x]$
$= \{(3 x+4 y)^2 +x^2\}(3 x+4 y+x)(3 x+y-x)$
$= \{(3 x+4 y)^2+x^2\}(4 x+4 y)(2 x+4 y)$
$=\{3 x+4 y)^2+x^2\} 4(x+y) 2(x+2 y)$
$=8\left\{(3 x+4 y)^2+x^2\right\}(x+y)(x+2 y)$
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