If a - b -c are real and distinct number - then the value of a - b3 - b - c3 - c - a3a - bb - cc - a is

The correct option is C 3 We know, #160; x ^ 3 + y ^ 3 + z ^ 3 =3xyz if x+y+z=0 Since #160; #160; ( a b ) #160; ^ 3 +( b c ) #160; ^ ...

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Standard XII

Mathematics

Permutations and Combinations

If a , b ,c...

Question

If $a, b, c$ are real and distinct number , then the value of $\frac{{(a - b)}^{3} + {(b - c)}^{3} + {(c - a)}^{3}}{(a - b) (b - c) (c - a)}$ is

1

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a b c

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3

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Solution

The correct option is C $3$
We know, $x^{3} + y^{3} + z^{3} = 3 x y z i f x + y + z = 0$
Since ${(a - b)}^{3} + {(b - c)}^{3} + {(c - a)}^{3}$ has terms $(a - b), (b - c) & (c - a)$ & they sum to zero
$∴ {(a - b)}^{3} + {(b - c)}^{3} + {(c - a)}^{3} = 3 (a - b) (b - c) (c - a)$
$∴ \frac{{(a - b)}^{3} + {(b - c)}^{3} + {(c - a)}^{3}}{(a - b) (b - c) (c - a)} = \frac{3 (a - b) (b - c) (c - a)}{(a - b) (b - c) (c - a)} = 3$

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If a,b,c are real and distinct number , then the value of (a−b)3+(b−c)3+(c−a)3(a−b)(b−c)(c−a) is

The correct option is C 3We know, x3+y3+z3=3xyzifx+y+z=0Since (a−b)3+(b−c)3+(c−a)3 has terms (a−b),(b−c)&(c−a) & they sum to zero∴(a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)∴(a−b)3+(b−c)3+(c−a)3(a−b)(b−c)(c−a)=3(a−b)(b−c)(c−a)(a−b)(b−c)(c−a)=3

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