Simplify the expression ${{\left( {{x}^{2}}+{{y}^{2}}-{{z}^{2}} \right)}^{2}}-{{\left( {{x}^{2}}-{{y}^{2}}+{{z}^{2}} \right)}^{2}}$

Hint: Use the algebraic identity ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac$. Hence expand the terms ${{\left( {{x}^{2}}+{{y}^{2}}-{{z}^{2}} \right)}^{2}}$ and ${{\left( {{x}^{2}}-{{y}^{2}}+{{z}^{2}} \right)}^{2}}$. Finally, combine the like terms and hence simplify the expression. Alternatively, use the algebraic identity $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$ and hence simplify the expression. Alternatively, write ${{\left( {{x}^{2}}+{{y}^{2}}-{{z}^{2}} \right)}^{2}}-{{\left( {{x}^{2}}-{{y}^{2}}+{{z}^{2}} \right)}^{2}}$ as ${{\left( {{x}^{2}}+\left( {{y}^{2}}-{{z}^{2}} \right) \right)}^{2}}-{{\left( {{x}^{2}}-\left( {{y}^{2}}-{{z}^{2}} \right) \right)}^{2}}$ and use the algebraic identity ${{\left( a+b \right)}^{2}}-{{\left( a-b \right)}^{2}}=4ab$

Complete step-by-step answer:
Let $S={{\left( {{x}^{2}}+{{y}^{2}}-{{z}^{2}} \right)}^{2}}-{{\left( {{x}^{2}}-{{y}^{2}}+{{z}^{2}} \right)}^{2}}={{S}_{1}}-{{S}_{2}}$, where ${{S}_{1}}={{\left( {{x}^{2}}+{{y}^{2}}-{{z}^{2}} \right)}^{2}}$ and ${{S}_{2}}={{\left( {{x}^{2}}-{{y}^{2}}+{{z}^{2}} \right)}^{2}}$
Now, we have
${{S}_{1}}={{\left( {{x}^{2}}+{{y}^{2}}-{{z}^{2}} \right)}^{2}}$
We know that ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac$
Put $a={{x}^{2}},b={{y}^{2}}$ and $c=-{{z}^{2}}$, we get
${{\left( {{x}^{2}}+{{y}^{2}}-{{z}^{2}} \right)}^{2}}={{\left( {{x}^{2}} \right)}^{2}}+{{\left( {{y}^{2}} \right)}^{2}}+{{\left( -{{z}^{2}} \right)}^{2}}+2\left( {{x}^{2}} \right)\left( {{y}^{2}} \right)+2\left( {{y}^{2}} \right)\left( -{{z}^{2}} \right)+2\left( {{x}^{2}} \right)\left( -{{z}^{2}} \right)$
Hence, we have
${{S}_{1}}={{x}^{4}}+{{y}^{4}}+{{z}^{4}}+2{{x}^{2}}{{y}^{2}}-2{{x}^{2}}{{z}^{2}}-2{{y}^{2}}{{z}^{2}}$
Also, we have
${{S}_{2}}={{\left( {{x}^{2}}-{{y}^{2}}+{{z}^{2}} \right)}^{2}}$
We know that ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac$
Put $a={{x}^{2}},b=-{{y}^{2}}$ and $c={{z}^{2}}$, we get
${{\left( {{x}^{2}}-{{y}^{2}}+{{z}^{2}} \right)}^{2}}={{\left( {{x}^{2}} \right)}^{2}}+{{\left( -{{y}^{2}} \right)}^{2}}+{{\left( {{z}^{2}} \right)}^{2}}+2\left( {{x}^{2}} \right)\left( -{{y}^{2}} \right)+2\left( -{{y}^{2}} \right)\left( {{z}^{2}} \right)+2\left( {{x}^{2}} \right)\left( {{z}^{2}} \right)$
Hence, we have
${{S}_{2}}={{x}^{4}}+{{y}^{4}}+{{z}^{4}}-2{{x}^{2}}{{y}^{2}}+2{{x}^{2}}{{z}^{2}}-2{{y}^{2}}{{z}^{2}}$
Hence substituting the values of ${{S}_{1}}$ and ${{S}_{2}}$ in the equation for S, we have
$S={{S}_{1}}-{{S}_{2}}={{x}^{4}}+{{y}^{4}}+{{z}^{4}}+2{{x}^{2}}{{y}^{2}}-2{{x}^{2}}{{z}^{2}}-2{{y}^{2}}{{z}^{2}}-\left( {{x}^{4}}+{{y}^{4}}+{{z}^{4}}-2{{x}^{2}}{{y}^{2}}+2{{x}^{2}}{{z}^{2}}-2{{y}^{2}}{{z}^{2}} \right)$
Simplifying by combining like terms, we get
$S=4{{x}^{2}}{{y}^{2}}-4{{x}^{2}}{{z}^{2}}$
Hence, we have
${{\left( {{x}^{2}}+{{y}^{2}}-{{z}^{2}} \right)}^{2}}-{{\left( {{x}^{2}}-{{y}^{2}}+{{z}^{2}} \right)}^{2}}=4{{x}^{2}}{{y}^{2}}-4{{x}^{2}}{{z}^{2}}$, which is the required simplified form of the expression.

Note: [1] Alternative solution:
We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
Taking $a={{x}^{2}}+{{y}^{2}}-{{z}^{2}}$ and $b={{x}^{2}}-{{y}^{2}}+{{z}^{2}}$ and using the above algebraic identity, we have
${{a}^{2}}-{{b}^{2}}=\left( {{x}^{2}}+{{y}^{2}}-{{z}^{2}}+{{x}^{2}}-{{y}^{2}}+{{z}^{2}} \right)\left( {{x}^{2}}+{{y}^{2}}-{{z}^{2}}-{{x}^{2}}+{{y}^{2}}-{{z}^{2}} \right)$
Hence on simplification, we have
${{\left( {{x}^{2}}+{{y}^{2}}-{{z}^{2}} \right)}^{2}}-{{\left( {{x}^{2}}-{{y}^{2}}+{{z}^{2}} \right)}^{2}}=\left( 2{{x}^{2}} \right)\left( 2{{y}^{2}}-2{{z}^{2}} \right)=4{{x}^{2}}{{y}^{2}}-4{{x}^{2}}{{z}^{2}}$, which is the same as obtained above.
[2] Alternative solution:
We have ${{\left( {{x}^{2}}+{{y}^{2}}-{{z}^{2}} \right)}^{2}}-{{\left( {{x}^{2}}-{{y}^{2}}+{{z}^{2}} \right)}^{2}}={{\left( {{x}^{2}}+\left( {{y}^{2}}-{{z}^{2}} \right) \right)}^{2}}-{{\left( {{x}^{2}}-\left( {{y}^{2}}-{{z}^{2}} \right) \right)}^{2}}$
We know that ${{\left( a+b \right)}^{2}}-{{\left( a-b \right)}^{2}}=4ab$
Put $a={{x}^{2}}$ and $b={{y}^{2}}-{{z}^{2}}$ and using the above algebraic identity, we get
${{\left( {{x}^{2}}+{{y}^{2}}-{{z}^{2}} \right)}^{2}}-{{\left( {{x}^{2}}-{{y}^{2}}+{{z}^{2}} \right)}^{2}}=4\left( {{x}^{2}} \right)\left( {{y}^{2}}-{{z}^{2}} \right)$
Using distributive law of multiplication over subtraction, we get
${{\left( {{x}^{2}}+{{y}^{2}}-{{z}^{2}} \right)}^{2}}-{{\left( {{x}^{2}}-{{y}^{2}}+{{z}^{2}} \right)}^{2}}=4{{x}^{2}}{{y}^{2}}-4{{x}^{2}}{{z}^{2}}$, which is the same as obtained above.