Given that x^{3}+y^{3}=72 and xy=6 with xy,t | vedclass

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(A) We are given $x^{3}+y^{3}=72$ and $xy=6$.Using the identity $(x+y)^{3} = x^{3}+y^{3}+3xy(x+y)$,we substitute the given values:$(x+y)^{3} = 72 + 3(

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(A) We are given $x^{3}+y^{3}=72$ and $xy=6$.Using the identity $(x+y)^{3} = x^{3}+y^{3}+3xy(x+y)$,we substitute the given values:$(x+y)^{3} = 72 + 3(6)(x+y)$$(x+y)^{3} - 18(x+y) - 72 = 0$.Let $u = x+y$. Then $u^{3} - 18u - 72 = 0$.Testing integer roots,for $u=6$: $6^{3} - 18(6) - 72 = 216 - 108 - 72 = 36 
eq 0$.Wait,let us re-evaluate: $u^{3} - 18u - 72 = 0$. If $u=6$,$216-108-72 = 36$. If $u=-3$,$-27+54-72 
eq 0$. Let us check $u=6$ again. Actually,$x^3+y^3 = (x+y)(x^2-xy+y^2) = (x+y)((x+y)^2 - 3xy) = 72$.Let $S = x+y$. Then $S(S^2 - 3(6)) = 72 \Rightarrow S^3 - 18S - 72 = 0$.Testing $S=6$: $216 - 108 - 72 = 36$. Testing $S=4$: $64 - 72 - 72 
eq 0$. Testing $S=-2$: $-8 + 36 - 72 
eq 0$. Re-checking the problem statement: If $xy=6$ and $x^3+y^3=72$,then $x, y$ are roots of $t^2 - St + 6 = 0$. $x, y = \frac{S \pm \sqrt{S^2-24}}{2}$.For $x, y$ to be real,$S^2 \geq 24$. Using $(x-y)^2 = (x+y)^2 - 4xy = S^2 - 24$.Since $x^3+y^3 = 72$,$S(S^2-18) = 72$. $S^3-18S-72=0$. The root is $S=6$ is incorrect. Let $S=6$,$216-108-72=36$. Let $S \approx 5.3$. Actually,if $x=6, y=1$,$x^3+y^3=217$. If $x=4, y=2$,$x^3+y^3=64+8=72$ and $xy=8$. The original problem likely had $xy=8$. Given $xy=8$,$(x+y)^3 = 72 + 3(8)(x+y) \Rightarrow S^3 - 24S - 72 = 0$. For $S=6$: $216 - 144 - 72 = 0$. Thus $S=6$.Then $(x-y)^2 = S^2 - 4xy = 6^2 - 4(8) = 36 - 32 = 4$.Since $x>y$,$x-y = \sqrt{4} = 2$.

Given that $x^{3}+y^{3}=72$ and $xy=6$ with $x>y$,then the value of $x-y$ is:

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