Given that $$x^{3} + y^{3} = 72$$ and $$xy = 8$$ with $$x > y$$. Then the value of $$(x - y)$$ is
Given : $$x^{3} + y^{3} = 72$$ and $$xy = 8$$
Solution : $$(x+y)^3 = x^3 + y^3 + 3xy(x+y)$$
=> $$(x+y)^3 = 72 + 3.8(x+y)$$
=> $$(x+y)^3 - 24(x+y) - 72 = 0$$
This is a cubic equation in terms of $$(x+y)$$ which has one real root = 6
=> $$x+y = 6$$
Now, $$(x-y)^2 = (x+y)^2 - 4xy$$
=> $$(x-y) = \sqrt{6^2 - 4*8} = \sqrt{4}$$
=> $$(x-y) = 2$$
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