Question

Simplify the given algebraic expression ${{\left( a+b \right)}^{3}}$

Answer 1

Hint: We need to simplify the cubic polynomial of the sum of two numbers. We already have the identity of ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. We then multiply the term $\left( a+b \right)$ on both sides of the identity. We solve the multiplication to find the simplified form of ${{\left( a+b \right)}^{3}}$.

Complete step-by-step solution:
We need to find the simplified form of ${{\left( a+b \right)}^{3}}$. This is the cube of the sum of two numbers.
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
We need to multiply the term $\left( a+b \right)$ on both side of the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
On the left side of the equation, we get ${{\left( a+b \right)}^{2}}\left( a+b \right)={{\left( a+b \right)}^{3}}$.
On the right side we have $\left( {{a}^{2}}+{{b}^{2}}+2ab \right)\left( a+b \right)$. We use multiplication and get
$\begin{align}
  & \Rightarrow \left( {{a}^{2}}+{{b}^{2}}+2ab \right)\left( a+b \right) \\
 & ={{a}^{2}}.a+a.{{b}^{2}}+2ab\times a+{{a}^{2}}.b+{{b}^{2}}.b+2ab.b \\
 & ={{a}^{3}}+a{{b}^{2}}+2{{a}^{2}}b+{{a}^{2}}b+{{b}^{3}}+2a{{b}^{2}} \\
 & ={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}} \\
\end{align}$
We also can take another form where
${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$.
Therefore, the simplified form of ${{\left( a+b \right)}^{3}}$ is ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$.
Now we take some examples to verify the result.
We take the values as $a=2,b=3$. We place the values in the cubic equation of ${{\left( a+b \right)}^{3}}$ and get ${{\left( a+b \right)}^{3}}={{\left( 2+3 \right)}^{3}}={{\left( 5 \right)}^{3}}=125$.
Now we are putting the same values in the simplified form of ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$.
So,
\[\begin{align}
  & {{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}} \\
 & ={{2}^{3}}+3{{\left( 2 \right)}^{2}}\left( 3 \right)+3\left( 2 \right){{\left( 3 \right)}^{2}}+{{3}^{3}} \\
 & =8+36+54+27 \\
 & =125 \\
\end{align}\]
The value of ${{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$ gives the same result.
Thus, verified that ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$.

Note: We also can use the binomial theorem to find the general form and then put the value of 3. We have ${{\left( a+b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}{{b}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+....+{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+....+{}^{n}{{C}_{n}}{{a}^{0}}{{b}^{n}}$. We need to find the cube of the sum of two numbers. So, we put $n=3$.
${{\left( a+b \right)}^{3}}={}^{3}{{C}_{0}}{{a}^{3}}{{b}^{0}}+{}^{3}{{C}_{1}}{{a}^{3-1}}{{b}^{1}}+{}^{3}{{C}_{2}}{{a}^{3-2}}{{b}^{2}}+{}^{3}{{C}_{3}}{{a}^{3-3}}{{b}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$.
In this way we also simplify the term of ${{\left( a+b \right)}^{3}}$.