If a^2 + b^2 = 99 and ab = 11, (a > 0, b > 0) then the value of (a^3 + b^3) is:

Question 56

If $$a^2 + b^2 = 99 and ab = 11, (a > 0, b > 0)$$ then the value of $$(a^3 + b^3)$$ is:

Solution

$$a^2 + b^2 = 99 and ab = 11$$

$$(a^3 + b^3)$$ = (a +b)$$(a^2 + b^2 - ab)$$

Hence we have to find a+b

$$a^2 + b^2 = 99 $$

$$a^2 + b^2 + 2ab= 99 + 22$$

$$(a + b)^2$$ = 121

$$(a + b)^2$$ = $$11^2$$

a + b =11

Hence,

$$(a^3 + b^3) =11 \times (88) = 968$$

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