Question

How do you factorize $14{x^6} - 45{x^3}{y^3} - 14{y^6}$?

Answer 1

Hint: As the given equation is quadratic in two variables, we will use the factorization method to find the factors of the given equation. . Now, factorize the given equation using splitting the middle term. Find two factors of -196 such that when they are added or subtracted, you will get -45. Then, take common and make factors. This will give you the answer.

Complete step-by-step answer:
We have been given an equation $14{x^6} - 45{x^3}{y^3} - 14{y^6}$.
We have to find the factors of the given equation.
The next step is to find two factors of -196 in such a way that when those factors are added or subtracted, we get -45. The two such factors can be 4 and -49.
Let us put them in the equation.
$ \Rightarrow 14{x^6} - 49{x^3}{y^3} + 4{x^3}{y^3} - 14{y^6}$
Now, take common from the terms,
$ \Rightarrow 7{x^3}\left( {2{x^3} - 7{y^3}} \right) + 2{y^3}\left( {2{x^3} - 7{y^3}} \right)$
Again, take commonly from the terms,
$ \Rightarrow \left( {7{x^3} + 2{y^3}} \right)\left( {2{x^3} - 7{y^3}} \right)$
So, this is the factorized form.

Hence, the two factors of the equation $14{x^6} - 45{x^3}{y^3} - 14{y^6}$ is $\left( {7{x^3} + 2{y^3}} \right)\left( {2{x^3} - 7{y^3}} \right)$.

Note:
As we know the form of quadratic equation in two variables is ${x^2} - \left( {\alpha + \beta } \right)xy + {y^2} = 0$ or ${x^2} - $(Sum of roots)$xy + $Product of roots $ = 0$. The factorization method uses the same concept. For the verification of splitting, we can check the brackets formed of factorization. If the two brackets formed after taking common parts from the first two terms and last two terms; these should be equal. If the two brackets formed are not equal, then splitting has gone wrong and we need to check the splitting step once again. Students make mistakes during taking common between the first two and last two terms. We need to take the common HCF of the first two and last two terms respectively, then we will get two brackets formed which is equal.