Question:
Simplify:
\((x^2 – y^2)^3 + (y^2 – z^2)^3 + (z^2 – x^2)^3\)
सरल कीजिए:
\((x^2 – y^2)^3 + (y^2 – z^2)^3 + (z^2 – x^2)^3\)
Options:
- \(3 (x – y)(y – z)(z – x) (x + y)(y + z)(z + x)\)
- \((x – y)(y – z)(z – x)\)
- \((x + y)(y + z)(z + x)\)
- \((x – y)^3\)
- \(3 (x – y)(y – z)(z – x) (x + y)(y + z)(z + x)\)
- \((x – y)(y – z)(z – x)\)
- \((x + y)(y + z)(z + x)\)
- \((x – y)^3\)
Correct Answer: a)
Explanation:
If \(a + b + c = 0\), then \(a^3 + b^3 + c^3 = 3abc\).
Here, \((x^2 – y^2) + (y^2 – z^2) + (z^2 – x^2) = 0\).
So the expression = \(3(x^2 – y^2)(y^2 – z^2)(z^2 – x^2)\)
\(= 3(x – y)(x + y)(y – z)(y + z)(z – x)(z + x)\).
यदि \(a + b + c = 0\) हो, तो \(a^3 + b^3 + c^3 = 3abc\) होता है।
यहाँ \(a = x^2 – y^2, b = y^2 – z^2, c = z^2 – x^2\) है, जिनका योग 0 है।
अतः व्यंजक = \(3 (x – y)(y – z)(z – x)(x + y)(y + z)(z + x)\)।