Question

How do you factor $100-81{{t}^{6}}$ ?

Answer 1

Hint: To factor the given expression $100-81{{t}^{6}}$, as you can see that 100 is the square of 10 and 81 is the square of 9 so write 100 and 81 as the square of 10 and 9 respectively. Similarly, we can write ${{t}^{6}}$ as ${{\left( {{t}^{3}} \right)}^{2}}$ in this expression. Then we are going to use the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$.

Complete step-by-step solution:
The expression given in the above problem which we are going to factorize is as follows:
$100-81{{t}^{6}}$
Now, writing 100 as the square of 10 and 81 as the square of 9 and ${{t}^{6}}$ as ${{\left( {{t}^{3}} \right)}^{2}}$ in the above expression and we get,
$\begin{align}
  & \Rightarrow 100-81{{t}^{6}} \\
 & ={{\left( 10 \right)}^{2}}-{{\left( 9{{t}^{3}} \right)}^{2}} \\
\end{align}$
To factorize further, we are going to use the algebraic identity which is equal to:
$\Rightarrow {{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$
Substituting the value of “a” and “b” as 10 and $9{{t}^{3}}$ in the above equation we get,
$\Rightarrow {{\left( 10 \right)}^{2}}-{{\left( 9{{t}^{3}} \right)}^{2}}=\left( 10-9{{t}^{3}} \right)\left( 10+9{{t}^{3}} \right)$
From the above factorization, we have got the factors as $\left( 10-9{{t}^{3}} \right)\left( 10+9{{t}^{3}} \right)$.

Note: In such problems where you have to find the factors, first of all, look at the terms of which we can write as square or cube of something then we see if we can use the algebraic identities. One of the identities, we have shown above. In the below, we have shown some of the algebraic identities which can be possibly useful in solving the problems:
$\begin{align}
  & {{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right); \\
 & {{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right) \\
\end{align}$
Now, you might think that in the above solution, we can use the above properties because you can see one of the cube terms in both the brackets.
This is the factors which we are getting in the above solution as follows:
$\left( 10-9{{t}^{3}} \right)\left( 10+9{{t}^{3}} \right)$
So, as you can see the cube of “t” in the above expression so we might use the cubic algebraic identities but we cannot because we only have cube of “t”, other terms are not in the form of some number to the power of three.