If a^2 + b^2 = 135 and ab = 7, (a > 0, b > 0) then the value of (a^3 - b^3) is:

Question 64

If $$a^2 + b^2 = 135 and ab = 7, (a > 0, b > 0)$$ then the value of $$(a^3 - b^3)$$ is:

Solution

$$a^2+b^2=135$$; ab =7
Now we know $$\left(a-b\right)^2=a^2+b^2-2ab$$
we get $$\left(a-b\right)^2=135-14=121$$
so a-b = 11
Now $$a^3-b^3=\ \left(a-b\right)\left(a^2+b^2+ab\right)$$
we get 11(135+7) = 11(142) =1562

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SSC CHSL 5 July 2019 Shift-1

If $$a^2 + b^2 = 135 and ab = 7, (a > 0, b > 0)$$ then the value of $$(a^3 - b^3)$$ is:

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