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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

Evaluate $\frac{0.53 \times 0.53 - 2 \times 0.53 \times 0.41 + 0.41 \times 0.41}{0.53 - 0.41}$

Simplify: $2 \div [2 + 2 \div \{2 + 2 \div (2 + 2 \div 3)\}]$

If $N = \frac{\sqrt{6} - \sqrt{5}}{\sqrt{6} + \sqrt{5}},$ then what is the value of $N + \frac{1}{N} ?$

What approximate value will come in place of the question mark $(?)$ in the given question? (You are not expected to calculate the exact value)
$71.98 \%$ of $1200 + 35.06 \%$ of $270 = ? \%$ of $600$

If $2x + \frac{2}{x} = 3$,then the value of $x^3 + \frac{1}{x^3} + 2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module