Question
Factorise:
$a^3(b - c)^3 + b^3(c - a)^3 + c^3(a - b)^3$
$a^3(b - c)^3 + b^3(c - a)^3 + c^3(a - b)^3$
✓
Answer
We have:
$a^3(b - c)^3 + b^3(c - a)^3 + c^3(a - b)^3 =$
$[a(b - c)]^3 + [a(b - c)]^3 + [b(c - a)]^3 + [c(a - b)]^3$
Put,
$a(b - c) = x, b(c - a) = y, c(a - b) = z$
Here,
$x + y + z = a(b - c) + b(c - a) + c(a - b)$
$= ab - ac + bc - ab - ab + ac - bc$
Thus,
We have:
$a^3(b - c)^3 + b^3(c - a)^3 + c^3(a - b)^3 = x^3 + y^3 + z^3$
$= 3xyz [When x + y + z = 0, x^3 + y^3 + z^3 = 3xyz]$
$= 3a(b - c)b(c - a)c(a -b)$
$= 3abc(a - b)(b - c)(c - a)$
$a^3(b - c)^3 + b^3(c - a)^3 + c^3(a - b)^3 =$
$[a(b - c)]^3 + [a(b - c)]^3 + [b(c - a)]^3 + [c(a - b)]^3$
Put,
$a(b - c) = x, b(c - a) = y, c(a - b) = z$
Here,
$x + y + z = a(b - c) + b(c - a) + c(a - b)$
$= ab - ac + bc - ab - ab + ac - bc$
Thus,
We have:
$a^3(b - c)^3 + b^3(c - a)^3 + c^3(a - b)^3 = x^3 + y^3 + z^3$
$= 3xyz [When x + y + z = 0, x^3 + y^3 + z^3 = 3xyz]$
$= 3a(b - c)b(c - a)c(a -b)$
$= 3abc(a - b)(b - c)(c - a)$
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