Question

expand (2x^2 +3)^4 and
(2x-1/x)^6 ​

Answer 1

Given:

$\bold{ (2x^2+3)^4 \ and (\frac{2x-1}{x})^6}$

To prove:

expand.

Solution:

Formula:

$\bold{(a+b)^2 =a^2+b^2+2ab}\\\\\bold{(a+b+c)^2= a^2+b^2+c^2+2ab+2bc+2ca}\\\\\bold{(a-b)^3=a^3-b^3-3ab(a-b)}\\\\\bold{(\frac{a}{b})^c =\frac{a^c}{b^c}}\\\\\bold{(a+b)^n=\sum_{k=0}^n \binom{n}{k} a^{n-k} b^k}$

$\Rightarrow \bold{ (2x^2+3)^4} =((2x^2+3)^2)^2$          

                      $= ((2x^2)^2+(3)^2+2\cdot 2x^2\cdot 3)^2\\\\= (4x^4+9+12x^2)^2\\\\= (4x^4)^2+(9)^2+(12x^2)^2+2 \cdot 4x^4 \cdot 9+ 2 \cdot 9 \cdot 12x^2+2\cdot 12x^2 \cdot 4x^2 \\\\= 16x^8+81+144x^4+72x^4 + 216 x^2+96x^6$

                     $= 16x^8+96x^6+216x^4+216x^2+81$

$\Rightarrow \bold{(\frac{2x-1}{x})^6}$

apply binomial theorem formula:                  

$(a+b)^n=\sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$

where,  $n!=1 \cdot 2 \cdot 3 ... n$

$a= - \frac{1}{x}\\ b= 2\\n=6$

put k = 1, 2, 3, 4, 5, 6

It will give:

$= 64 - \frac{192}{x} + \frac{240}{x^{2}} - \frac{160}{x^{3}} + \frac{60}{x^{4}} - \frac{12}{x^{5}} + \frac{1}{x^{6}}$

The Expanded form of  $\bold{(2x^2+3)^4 = 16x^8+96x^6+216x^4+216x^2+81 }$

and $\bold{ \left(2 - \frac{1}{x}\right)^{6}=64 - \frac{192}{x}+\frac{240}{x^{2}}-\frac{160}{x^{3}} + \frac{60}{x^{4}} - \frac{12}{x^{5}} + \frac{1}{x^{6}}}$.