If a2-b2-c2-60- a b c-4- 1-a-1-b-1-c-3- then the values of a-b-c- is-are

The correct options are A 6 B 6(a+b+c)^2=a^2+b^2+c^2+2abc(1a+1b+1c) ⇒[a+( b)+c]^2=a^2+( b)^2+c^2+2a( b)c[1a+( 1b)+1c] ⇒[a+( b)+c]^2=60+2( 4)(3) ⇒(a b+c)^2=36 ⇒ ...

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Standard XII

Mathematics

Cartesian Product

If a2+b2+c2=6...

Question

If $a^{2} + b^{2} + c^{2} = 60, a b c = 4,$ $\frac{1}{a} - \frac{1}{b} + \frac{1}{c} = 3$ , then the value(s) of $a - b + c$ , is/are

6

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\sqrt{48}

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Solution

The correct options are
A $6$
B $- 6$
$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b c (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$
$\Rightarrow [a + (- b) + c]^{2} = a^{2} + (- b)^{2} + c^{2} + 2 a (- b) c [\frac{1}{a} + (- \frac{1}{b}) + \frac{1}{c}]$
$\Rightarrow [a + (- b) + c]^{2} = 60 + 2 (- 4) (3)$
$\Rightarrow (a - b + c)^{2} = 36$
$\Rightarrow a - b + c = \pm 6$

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Cartesian Product

Standard XII Mathematics

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If a2+b2+c2=60, abc=4, 1a−1b+1c=3, then the value(s) of a−b+c, is/are

The correct options are A 6 B −6(a+b+c)2=a2+b2+c2+2abc(1a+1b+1c) ⇒[a+(−b)+c]2=a2+(−b)2+c2+2a(−b)c[1a+(−1b)+1c] ⇒[a+(−b)+c]2=60+2(−4)(3) ⇒(a−b+c)2=36 ⇒a−b+c=±6

If $a^{2} + b^{2} + c^{2} = 60, a b c = 4,$ $\frac{1}{a} - \frac{1}{b} + \frac{1}{c} = 3$ , then the value(s) of $a - b + c$ , is/are