Question

acube breketemay b minus C ka whole cube plus b cube bracket Mec- a ka whole cube plus c cube bracket a
minus b ka whole cube factorise

Answer 1

${a}^{3} {(b - c)}^{3} + {b}^{3} {(c - a)}^{3} + {c}^{3} {(a - b)}^{3} \\ a(b - c) + b(c - a) + c(a - b) \\ = ab - ac + bc - ab + ab - bc \\ = 0 \\ so \\ {(a(b - c)})^{3} + {(b(c - a))}^{3} + {(c(a - b))}^{3} = 3(ab - ac)(bc - ab)(ab - bc).$

Answer 2

The answer to the above question is $3abc(a-b)(b-c)(c-a)$.

Given:

$a^3(b-c)^3+b^3(c-a)^3+c^3(a-b)^3$

To Find:

The solution or to simplify the above-given equation.

Solution:

We know that if $x+y+z=0$, then $x^3+y^3+z^3=3xyz$.

Let,

$x=a(b-c)\\y=b(c-a)\\z=c(a-b)$

Therefore,

$x+y+z=a(b-c)+b(c-a)+c(a-b)\\x+y+z=ab-ac+bc-ba+ca-cb\\x+y+z=0$

Hence, $x^3+y^3+z^3=3xyz$

Therefore,

$a^3(b-c)^3+b^3(c-a)^3+c^3(a-b)^3$  $=3((a(b-c))(b(c-a))(c(a-b)))$

$a^3(b-c)^3+b^3(c-a)^3+c^3(a-b)^3 = 3abc(a-b)(b-c)(c-a)$

Hence, $3abc(a-b)(b-c)(c-a)$ is the required solution.

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