Factorise $(a^{2} - b^{2}) c + (b^{2} - c^{2}) a$

(a + c) (b^{2} + a c)

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(a - c) (b^{2} + c)

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(a - c) (b^{2} + a)

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(a - c) (b^{2} + a c)

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Solution

The correct option is D $(a - c) (b^{2} + a c)$
$F a c t o r i z i n g {(a^{2} - b}^{2}) c + {(b}^{2} - c^{2}) a, = a^{2} c - b^{2} c + a b^{2} - a c^{2}, r e a r r a n g e a n d, = b^{2} (a - c) + a c (a - c) = (a - c) (a c + b^{2})$ \\

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Similar questions

If $a, b, c$ are in G.P., then ___.

|\begin{matrix} 1 & b + c & b^{2} + c^{2} \\ 1 & c + a & c^{2} + a^{2} \\ 1 & a + b & a^{2} + b^{2} \end{matrix}| = (a - b) (b - c) (c - a)

Δ A B C, \frac{a}{b^{2} - c^{2}} + \frac{c}{b^{2} - a^{2}} = 0

B =

P = a^{2} + a b^{2} - a^{2} c - a c^{2}

Q = b^{2} + b c^{2} - a^{2} b - a b^{2}

R = a^{2} c + c^{2} a - c^{2} b - c b^{2}

P x^{2} + Q x + R = 0

(\frac{x^{a}}{x^{- b}})^{a^{2} - a b + b^{2}} \times (\frac{x^{b}}{x^{- c}})^{b^{2} - b c + c^{2}} \times (\frac{x^{c}}{x^{- a}})^{c^{2} - c a + a^{2}} =

x^{2 (a^{3} + b^{3} + c^{3})}

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