Out of the given responses,one of the factors | vedclass

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(A) Let $x = a^{2}-b^{2}$,$y = b^{2}-c^{2}$,and $z = c^{2}-a^{2}$.Then,$x+y+z = (a^{2}-b^{2}) + (b^{2}-c^{2}) + (c^{2}-a^{2}) = 0$.We know the algebra

Vedclass · Accepted Answer

$(a+b)(a-b)$

Vedclass · Answer

(A) Let $x = a^{2}-b^{2}$,$y = b^{2}-c^{2}$,and $z = c^{2}-a^{2}$.Then,$x+y+z = (a^{2}-b^{2}) + (b^{2}-c^{2}) + (c^{2}-a^{2}) = 0$.We know the algebraic identity: If $x+y+z = 0$,then $x^{3}+y^{3}+z^{3} = 3xyz$.Substituting the values,we get:$(a^{2}-b^{2})^{3}+(b^{2}-c^{2})^{3}+(c^{2}-a^{2})^{3} = 3(a^{2}-b^{2})(b^{2}-c^{2})(c^{2}-a^{2})$.Expanding the terms using the difference of squares formula $a^{2}-b^{2} = (a-b)(a+b)$:$= 3(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)$.Thus,$(a+b)(a-b)$ is one of the factors.

Out of the given responses,one of the factors of $(a^{2}-b^{2})^{3}+(b^{2}-c^{2})^{3}+(c^{2}-a^{2})^{3}$ is:

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