Question

Simplify the algebraic expression: ${{x}^{4}}+{{x}^{2}}{{y}^{2}}+{{y}^{4}}$.

Answer 1

Hint: We first equate the equation ${{x}^{4}}+{{x}^{2}}{{y}^{2}}+{{y}^{4}}$ with 0. We then form a square for the left side of the new equation. Then we take the square root on both sides of the equation. From that we form the factorisation of ${{x}^{4}}+{{x}^{2}}{{y}^{2}}+{{y}^{4}}$.

Complete step-by-step solution:
We need to find the simplified form of the given equation ${{x}^{4}}+{{x}^{2}}{{y}^{2}}+{{y}^{4}}$. We equate with 0 and get ${{x}^{4}}+{{x}^{2}}{{y}^{2}}+{{y}^{4}}=0$.
We use the identity ${{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}$.
\[\begin{align}
  & {{x}^{4}}+{{x}^{2}}{{y}^{2}}+{{y}^{4}}=0 \\
 & \Rightarrow {{x}^{4}}+2{{x}^{2}}{{y}^{2}}+{{y}^{4}}={{x}^{2}}{{y}^{2}} \\
\end{align}\]
We interchanged the numbers for $a={{x}^{2}},b={{y}^{2}}$.
\[\begin{align}
  & {{x}^{4}}+2{{x}^{2}}{{y}^{2}}+{{y}^{4}}={{x}^{2}}{{y}^{2}} \\
 & \Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}={{\left( xy \right)}^{2}} \\
 & \Rightarrow {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-{{\left( xy \right)}^{2}}=0 \\
\end{align}\]
For the factorisation of the given quadratic polynomial \[{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-{{\left( xy \right)}^{2}}=0\], we apply the factorisation identity of difference of two squares as ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
We put the value of $a=\left( {{x}^{2}}+{{y}^{2}} \right);b=\left( xy \right)$.
Factorisation of the polynomial gives us \[{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-{{\left( xy \right)}^{2}}=\left( {{x}^{2}}+{{y}^{2}}+xy \right)\left( {{x}^{2}}+{{y}^{2}}-xy \right)\].
These two multiplied linear polynomials can’t be broken any more.
Therefore, the final factorisation of ${{x}^{4}}+{{x}^{2}}{{y}^{2}}+{{y}^{4}}$ is \[\left( {{x}^{2}}+{{y}^{2}}+xy \right)\left( {{x}^{2}}+{{y}^{2}}-xy \right)\].

Note: The highest power of the variable or the degree of a polynomial decides the number of roots or the solution of that polynomial. Quadratic equations have two roots. Cubic polynomials have three. It can be both real and imaginary roots.
In the given polynomial the values will be dependent on the second variable.