Question

How do you use binomial theorem to expand and simplify the expression ${\left( {{x^2} + {y^2}} \right)^6}$ ?

Answer 1

Hint : The given question requires us to find the binomial expansion of the given binomial expression. We can find the required binomial expansion by using the Binomial theorem. Binomial theorem helps us to expand the powers of binomial expressions easily and can be used to solve the given problem. We must know the formulae of combinations and factorials for solving such questions using the binomial theorem.

Complete step-by-step answer :
We have to find the binomial expansion of ${\left( {{x^2} + {y^2}} \right)^6}$ . So, using the binomial theorem, the binomial expansion of ${\left( {x + y} \right)^n}$ is $\sum\nolimits_{r = 0}^n {\left( {^n{C_r}} \right){{\left( x \right)}^{n - r}}{{\left( y \right)}^r}}$
So, the binomial expansion of ${\left( {{x^2} + {y^2}} \right)^6}$ is
$\sum\nolimits_{r = 0}^6 {\left( {^6{C_r}} \right){{\left( {{x^2}} \right)}^{6 - r}}{{\left( {{y^2}} \right)}^r}}$ .
Now, we have to expand the expression
$\sum\nolimits_{r = 0}^6 {\left( {^6{C_r}} \right){{\left( {{x^2}} \right)}^{6 - r}}{{\left( {{y^2}} \right)}^r}}$ and we are done with the simplification of expression ${\left( {{x^2} + {y^2}} \right)^6}$ . So, we have,
$= \left( {^6{C_0}} \right){\left( {{x^2}} \right)^6}{\left( {{y^2}} \right)^0} + \left( {^6{C_1}} \right){\left( {{x^2}} \right)^5}{\left( {{y^2}} \right)^1} + \left( {^6{C_2}} \right){\left( {{x^2}} \right)^4}{\left( {{y^2}} \right)^2} + \left( {^6{C_3}} \right){\left( {{x^2}} \right)^3}{\left( {{y^2}} \right)^3} + \left( {^6{C_4}} \right){\left( {{x^2}} \right)^2}{\left( {{y^2}} \right)^4} + \left( {^6{C_5}} \right){\left( {{x^2}} \right)^1}{\left( {{y^2}} \right)^5} + \left( {^6{C_6}} \right){\left( {{x^2}} \right)^0}{\left( {{y^2}} \right)^6}$
Simplifying the expression further, we get,
$= \left( {^6{C_0}} \right){x^{12}} + \left( {^6{C_1}} \right){x^{10}}{y^2} + \left( {^6{C_2}} \right){x^8}{y^4} + \left( {^6{C_3}} \right){x^6}{y^6} + \left( {^6{C_4}} \right){x^4}{y^8} + \left( {^6{C_5}} \right){x^2}{y^{10}} + \left( {^6{C_6}} \right){y^{12}}$
Computing and substituting the values of combinations using the formula $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we get,
$= \left( 1 \right){x^{12}} + \left( 6 \right){x^{10}}{y^2} + \left( {15} \right){x^8}{y^4} + \left( {20} \right){x^6}{y^6} + \left( {15} \right){x^4}{y^8} + \left( 6 \right){x^2}{y^{10}} + \left( 1 \right){y^{12}}$
Simplifying the expressions further, we get,
$= {x^{12}} + 6{x^{10}}{y^2} + 15{x^8}{y^4} + 20{x^6}{y^6} + 15{x^4}{y^8} + 6{x^2}{y^{10}} + {y^{12}}$
Hence, the simplification of the expression ${\left( {{x^2} + {y^2}} \right)^6}$ using the binomial theorem is ${x^{12}} + 6{x^{10}}{y^2} + 15{x^8}{y^4} + 20{x^6}{y^6} + 15{x^4}{y^8} + 6{x^2}{y^{10}} + {y^{12}}$ .
So, the correct answer is “ ${x^{12}} + 6{x^{10}}{y^2} + 15{x^8}{y^4} + 20{x^6}{y^6} + 15{x^4}{y^8} + 6{x^2}{y^{10}} + {y^{12}}$ ”.

Note : The given problem can be solved by various methods. The easiest way is to apply the concepts of Binomial theorem as it is very effective in finding the binomial expansion. One should be careful while computing the combinations as it can be a tedious task as it involves cumbersome calculations. Also, algebraic simplifications should be handled with utmost care.