Find (a+b)^{4}-(a-b)^{4}. Hence,evaluate (sqr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Using the Binomial Theorem,the expansions are:$(a+b)^4 = \binom{4}{0}a^4 + \binom{4}{1}a^3b + \binom{4}{2}a^2b^2 + \binom{4}{3}ab^3 + \binom{4}{4}

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) Using the Binomial Theorem,the expansions are:$(a+b)^4 = \binom{4}{0}a^4 + \binom{4}{1}a^3b + \binom{4}{2}a^2b^2 + \binom{4}{3}ab^3 + \binom{4}{4}b^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$$(a-b)^4 = \binom{4}{0}a^4 - \binom{4}{1}a^3b + \binom{4}{2}a^2b^2 - \binom{4}{3}ab^3 + \binom{4}{4}b^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4$Subtracting the two expressions:$(a+b)^4 - (a-b)^4 = (a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4) - (a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4)$$= 8a^3b + 8ab^3 = 8ab(a^2 + b^2)$Now,substitute $a = \sqrt{3}$ and $b = \sqrt{2}$:$8(\sqrt{3})(\sqrt{2})((\sqrt{3})^2 + (\sqrt{2})^2) = 8\sqrt{6}(3 + 2) = 8\sqrt{6}(5) = 40\sqrt{6}$.

Find $(a+b)^{4}-(a-b)^{4}$. Hence,evaluate $(\sqrt{3}+\sqrt{2})^{4}-(\sqrt{3}-\sqrt{2})^{4}$. (in $\sqrt{6}$)

Similar Questions

The coefficient of $x^6$ in the power series expansion of $\frac{x^4-12x^2+7}{(x^2+1)^3}$ is

Let $(1 - 2x + 3x^2)^{10} = a_0 + a_1x + a_2x^2 + \dots + a_n x^n$,where $a_n \neq 0$. Then the arithmetic mean of $a_0, a_1, a_2, \dots, a_n$ is

If $|x|$ is so small that all terms containing $x^2$ and higher powers of $x$ can be neglected,then the approximate value of $\frac{(3-5x)^{1/2}}{(5-3x)^2}$,when $x=\frac{1}{\sqrt{363}}$,is

Expand the expression $\left(x+\frac{1}{x}\right)^{6}$.

The coefficient of $x^n$ in the expansion of $(1 + x + x^2 + ....)^{-n}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module