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Standard VIII

Mathematics

Expansion of (x+y+z)^2

If a,b > 0 ...

Question

If $a, b > 0$ satisfy $a^{3} + b^{3} = a - b$ then

a^{2} + b^{2} = 1

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a^{2} + a b + b^{2} < 1

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a^{2} + b^{2} > 1

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a^{2} - b^{2} = 1

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Solution

The correct option is A $a^{2} + a b + b^{2} < 1$
Given $a, b > 0$ , then we can write $(a + b) > (a - b) . . . . . (1)$
Also, $a - b = a^{3} + b^{3} . . . . (2)$
Putting the value of $a - b$ from equation (2) in the inequation (1)
$(a + b) > a^{3} + b^{3} \Rightarrow (a + b) > (a + b) (a^{2} - a b + b^{2})$
$\Rightarrow a^{2} - a b + b^{2} < 1$

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