If a- b- c be distinct real numbers such that displaystyle a-2-b- b-2-c- c-2-a- then displaystyle left - a-b right -left - b-c right -left - c-a right - is equal to01-1none of these

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Toppr Admin · Accepted Answer

Given-xA0-  a2-x2212-b-b2-x2212-c-c2-x2212-a-x21D2-a2-x2212-b2-b-x2212-c-b2-x2212-c2-c-x2212-a-c2-x2212-a2-a-x2212-b-x21D2-a-b-b-x2212-ca-x2212-b-b-c-c-x2212-ab-x2212-c-c-a-a-x2212-bc-x2212-a-x2234-a-b-b-c-c-a-1

If $a, b, c$ be distinct real numbers such that $a^{2} - b = b^{2} - c = c^{2} - a$ , then $(a + b) (b + c) (c + a)$ , is equal to
$0$
$1$
$- 1$
none of these

Given, $a^{2} - b = b^{2} - c = c^{2} - a$
$\Rightarrow a^{2} - b^{2} = b - c, b^{2} - c^{2} = c - a, c^{2} - a^{2} = a - b$
$\Rightarrow a + b = \frac{b - c}{a - b}, b + c = \frac{c - a}{b - c}, c + a = \frac{a - b}{c - a}$
$∴ (a + b) (b + c) (c + a) = 1$

If a,b,c be distinct real numbers such that a2−b=b2−c=c2−a, then (a+b)(b+c)(c+a), is equal to01−1none of these

Given, a2−b=b2−c=c2−a⇒a2−b2=b−c,b2−c2=c−a,c2−a2=a−b⇒a+b=b−ca−b,b+c=c−ab−c,c+a=a−bc−a∴(a+b)(b+c)(c+a)=1

If $a, b, c$ be distinct real numbers such that $a^{2} - b = b^{2} - c = c^{2} - a$ , then $(a + b) (b + c) (c + a)$ , is equal to
$0$
$1$
$- 1$
none of these

Given, $a^{2} - b = b^{2} - c = c^{2} - a$
$\Rightarrow a^{2} - b^{2} = b - c, b^{2} - c^{2} = c - a, c^{2} - a^{2} = a - b$
$\Rightarrow a + b = \frac{b - c}{a - b}, b + c = \frac{c - a}{b - c}, c + a = \frac{a - b}{c - a}$
$∴ (a + b) (b + c) (c + a) = 1$