# Simplify each of the following expressions:$(x+y+z)^{2}+\left(x+\frac{y}{2}+\frac{z}{3}\right)^{2}-\left(\frac{x}{2}+\frac{y}{3}+\frac{z}{4}\right)^{2}$

Given:

$(x+y+z)^{2}+\left(x+\frac{y}{2}+\frac{z}{3}\right)^{2}-\left(\frac{x}{2}+\frac{y}{3}+\frac{z}{4}\right)^{2}$

To do:

We have to simplify $(x+y+z)^{2}+\left(x+\frac{y}{2}+\frac{z}{3}\right)^{2}-\left(\frac{x}{2}+\frac{y}{3}+\frac{z}{4}\right)^{2}$.

Solution:

We know that,

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Therefore,

$(x+y+z)^{2}+(x+\frac{y}{2}+\frac{z}{3})^{2}-(\frac{x}{2}+\frac{y}{3}+\frac{z}{4})^{2}=(x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x)+(x^{2}+\frac{y^{2}}{4}+\frac{z^{2}}{9}+\frac{2 x y}{2} \times \frac{2y}{2} \times \frac{z}{3}+\frac{2 z}{3} \times x)-(\frac{x^{2}}{4}+\frac{y^{2}}{9}+\frac{z^{2}}{16}+\frac{2 x}{2}\times \frac{y}{3}+\frac{2 y}{3} \times \frac{z}{4}+\frac{2 z}{4} \times \frac{x}{2})$

$=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x+x^{2}+\frac{y^{2}}{4}+\frac{z^{2}}{9}+x y+\frac{y z}{3}+\frac{2 z x}{3}-\frac{x^{2}}{4}-\frac{y^{2}}{9}-\frac{z^{2}}{16}-\frac{x y}{3}-\frac{y z}{6}-\frac{x z}{4}$

$=x^{2}+x^{2}-\frac{x^{2}}{4}+y^{2}+\frac{y^{2}}{4}-\frac{y^{2}}{9}+z^{2}+\frac{z^{2}}{9}-\frac{z^{2}}{16}+2 x y+x y-\frac{x y}{3}+2 y z+\frac{y z}{3}-\frac{y z}{6}+2 z x+\frac{2 z x}{3}-\frac{z x}{4}$

$=\frac{4 x^{2}+4 x^{2}-x^{2}}{4}+\frac{36 y^{2}+9 y^{2}-4 y^{2}}{36}+\frac{144 z^{2}+16 z^{2}-9 z^{2}}{144}+\frac{6 x y+3 x y-x y}{3}+\frac{12 y z+2 y z-y z}{6}+\frac{24 z x+8 z x-3 z x}{12}$

$=\frac{7 x^{2}}{4}+\frac{41 y^{2}}{36}+\frac{151 z^{2}}{14}+\frac{8}{3} x y+\frac{13}{6} y z+\frac{29}{12} z x$

Hence, $(x+y+z)^{2}+(x+\frac{y}{2}+\frac{z}{3})^{2}-(\frac{x}{2}+\frac{y}{3}+\frac{z}{4})^{2}=\frac{7 x^{2}}{4}+\frac{41 y^{2}}{36}+\frac{151 z^{2}}{14}+\frac{8}{3} x y+\frac{13}{6} y z+\frac{29}{12} z x$.

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Updated on: 10-Oct-2022

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