If x+y=2, then the value of x3+6xy+y3-8 is

# If $x+y=2$, then the value of ${x}^{3}+6\mathit{xy}+{y}^{3}-8$ is

1. A
1
2. B
3
3. C
0
4. D
5

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### Solution:

It is given ${x}^{3}+6\mathit{xy}+{y}^{3}-8$.
Simplifying $x+y=2$,
$⟹x=2-y$
Substituting $x=2-y$ in ${x}^{3}+6\mathit{xy}+{y}^{3}-8$, we get,
${x}^{3}+6\mathit{xy}+{y}^{3}-8={\left(2-y\right)}^{3}+6\left(2-y\right)\left(y\right)+{y}^{3}-8$
Applying algebraic identity ${\left(x-y\right)}^{3}={x}^{3}-{y}^{3}-3\mathit{xy}\left(x-y\right)$, we get,
$⇒{2}^{3}-{y}^{3}-3\left(2\right)\left(y\right)\left(2-y\right)+\left(12-6y\right)y+{y}^{3}-8$ $⇒8-{y}^{3}-12y+6{y}^{2}+12y-6{y}^{2}+{y}^{3}-8$ $⇒8-8$
$⇒0$
Therefore, the value of ${x}^{3}+6\mathit{xy}+{y}^{3}-8$ is 0.
Hence, option 3 is  correct.

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