Question
Factorize:
$a(a + b)^3 - 3a^2b(a + b)$
$a(a + b)^3 - 3a^2b(a + b)$
✓
Answer
$a(a+b)^3-3 a^2 b(a+b)$
Taking $(a+b)$ common in the two terms
$=(a+b)\left\{a(a+b)^2-3 a^2 b\right\}$
Now, using $(a+b)^2=a^2+b^2+2 a b$
$=(a+b)\left\{a\left(a^2+b^2+2 a b\right)-3 a^2 b\right\}$
$=(a+b)\left\{a^3+a b^2+2 a^2 b-3 a^2 b\right\}$
$=(a+b)\left\{a^3+a b^2-a^2 b\right\}$
$=(a+b) p\left\{a^2+b^2-a b\right\}$
$=p(a+b)\left(a^2+b^2-a b\right)$
$\therefore a(a+b)^3-3 a^2 b(a+b)$
$=a(a+b)\left(a^2+b^2-a b\right)$
Taking $(a+b)$ common in the two terms
$=(a+b)\left\{a(a+b)^2-3 a^2 b\right\}$
Now, using $(a+b)^2=a^2+b^2+2 a b$
$=(a+b)\left\{a\left(a^2+b^2+2 a b\right)-3 a^2 b\right\}$
$=(a+b)\left\{a^3+a b^2+2 a^2 b-3 a^2 b\right\}$
$=(a+b)\left\{a^3+a b^2-a^2 b\right\}$
$=(a+b) p\left\{a^2+b^2-a b\right\}$
$=p(a+b)\left(a^2+b^2-a b\right)$
$\therefore a(a+b)^3-3 a^2 b(a+b)$
$=a(a+b)\left(a^2+b^2-a b\right)$
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