If a, b >0satisfy $a^{3} + b^{3} = a + b$ 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 that:
$a^{3} + b^{3} = (a + b)$
As we know the identity
$a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$
$(a + b) = (a + b) (a^{2} - a b + b^{2})$
Then
$(a^{2} - a b + b^{2}) = 1$

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If a, b >0satisfy a3+b3=a+b then

The correct option is A (a2−ab+b2)=1Given that:a3+b3=(a+b)As we know the identitya3+b3=(a+b)(a2−ab+b2)(a+b)=(a+b)(a2−ab+b2)Then(a2−ab+b2)=1

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

The correct option is A $(a^{2} - a b + b^{2}) = 1$
Given that:
$a^{3} + b^{3} = (a + b)$
As we know the identity
$a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$
$(a + b) = (a + b) (a^{2} - a b + b^{2})$
Then
$(a^{2} - a b + b^{2}) = 1$