Question

How do you divide $\dfrac{{{y}^{3}}+216}{y+6}$?

Answer 1

Hint: We first take the factorisation of the given polynomial ${{y}^{3}}+216$ in the numerator according to the identity ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$. The division cancels out the polynomial $\left( y+6 \right)$ in the factorisation and gives $\left( {{y}^{2}}-6y+36 \right)$ as the quotient. We also verify the result with the long division.

Complete step by step answer:
We have been given a fraction whose denominator and numerator both are polynomial.
We need to find the value of the quotient.
The numerator is a cubic expression. It’s a sum of two cube numbers. We factorise the given sum of the cubes according to the identity
${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$.
We have ${{y}^{3}}+216$ in the numerator which gives ${{y}^{3}}+216={{y}^{3}}+{{6}^{3}}$. We put the values $a=y;b=6$.
We get ${{y}^{3}}+{{6}^{3}}=\left( y+6 \right)\left( {{y}^{2}}-6y+36 \right)$.
We can see the term ${{y}^{3}}+216$ is a multiplication of two polynomials $\left( y+6 \right)$ and $\left( {{y}^{2}}-6y+36 \right)$.
So, when ${{y}^{3}}+216$ is divided by $\left( y+6 \right)$, the multiplication of two polynomials $\left( y+6 \right)$ and $\left( {{y}^{2}}-6y+36 \right)$ gets divided by $\left( y+6 \right)$. In result we get $\left( {{y}^{2}}-6y+36 \right)$ as the quotient.
Therefore, \[\dfrac{{{y}^{3}}+216}{y+6}=\dfrac{\left( y+6 \right)\left( {{y}^{2}}-6y+{{6}^{2}} \right)}{\left( y+6 \right)}=\left( {{y}^{2}}-6y+36 \right)\].
The result of the division $\dfrac{{{y}^{3}}+216}{y+6}$ is $\left( {{y}^{2}}-6y+36 \right)$.

Note: We can also solve the division in the long division method. We first equate the highest power of the divisor and the dividend. We get ${{y}^{2}}$ as a result. Then we have $-6{{y}^{2}}$ and the rest $216$. We try to with $-6{{y}^{2}}$. We get $-6y$ as the next term. Then we have terms $36y+216$ as the remaining of the equation to equate. We multiply 36 with $\left( y+6 \right)$ to complete the division. The quotients are the multipliers which give . Thus, the result is verified.
y+6 $\overset{{\underline{{{y}^{2}}-6y+36}}}{{\left|{\begin{align}
  & {{y}^{3}}+216 \\
 & \underline{{{y}^{3}}+6{{y}^{2}}} \\
 & -6{{y}^{2}}+216 \\
 & \underline{-6{{y}^{2}}-36y} \\
 & 36y+216 \\
 & \underline{36y+216} \\
 & 0 \\
\end{align}}\right.}}$