Question

pls answer it
class - 9 maths question
i will surely mark u r answer as brainlist​

Answer 1

solution

$= \frac{(a {}^{2} - b {}^{2}) {}^{3} + (b {}^{2} - c {}^{2}) {}^{3} + (c {}^{2} - a {}^{2}) {}^{3} }{(a - b) {}^{3} + (b - c) {}^{3} + (c - a) {}^{23} } \\ = \frac{3(a {}^{2} - b {}^{2} )(b {}^{2} - c {}^{2})(c {}^{2} - a {}^{2} )}{3(a - b)(b - c)(c - a)} \\ = \frac{3(a - b)(a + b)(b - c)(b + c)(c - a)(c + a)}{3(a - b)(b - c)(c - a)} \\ = (a + b)(b + c)(c + a)$

______________________________________

now..we know that ....

x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)

now if (x+y+z)=0

then...

x³+y³+z³=3xyz

here in this problem

(a²-b²)+(b²-c²)+(c²-a²)=0

and

(a-b)+(b-c)+(c-a)=0

.

.

.

.

. Hope this helps you......