If x+y =12 and xy = 27, x > y, then the value of (x^3 - y^3) is:

Question 62

If x+y =12 and xy = 27, x > y, then the value of $$(x^3 - y^3)$$ is:

Solution

Given that,

x+y =12 and xy = 27

We know that $$(x-y)=\sqrt{(x+y)^2-4xy}$$

$$\Rightarrow (x-y)=\sqrt{12^2-4\times 27}=\sqrt{144-108}$$

$$\Rightarrow (x-y)=\sqrt{36}=6$$

Now, We know that $$(x-y)^3=x^3-y^3-3xy(x-y)$$

$$\Rightarrow 6^3=x^3-y^3-3\times 27\times(6)$$

$$\Rightarrow 216=x^3-y^3-486$$

$$\Rightarrow x^3-y^3=486+216=702$$

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