Binomial Theorem — MATHS STD 11 Science — Question

$(\text{a}+\text{b})^4-(\text{a}-\text{b})^4$

$=\big[{^4\text{C}}_0\text{a}^4\text{b}^0+{^4\text{C}}_1\text{a}^3\text{b}^1+{^4\text{C}}_2\text{a}^2\text{b}^2{\text{C}}_3\text{a}^1\text{b}^3+{^4\text{C}}_4\text{a}^0\text{b}^4\big]\\-\big[{^4\text{C}}_0\text{a}^4\text{b}^0+{^4\text{C}}_1\text{a}^3\text{b}^1+{^4\text{C}}_2\text{a}^2\text{b}^2-{^4\text{C}}_3\text{a}^1\text{b}^3+{^4\text{C}}_4\text{a}^0\text{b}^4\big]$

$=\big[{^4\text{C}}_0\text{a}^4(-\text{b})^0+{^4\text{C}}_1\text{a}^3(-\text{b})^1+{^4\text{C}}_2\text{a}^2(-\text{b})+{^4\text{C}}_3\text{a}^1(-\text{b})^3+{^4\text{C}}_4\text{a}^0(-\text{b})^4\big]\\-\big[{^4\text{C}}_0\text{a}^4(-\text{b})^0+{^4\text{C}}_1\text{a}^3(-\text{b})^1+{^4\text{C}}_2\text{a}^2(-\text{b})^2+{^4\text{C}}_3\text{a}^1(-\text{b})^3+{^4\text{C}}_4\text{a}^0(-\text{b})^4\big]$

$=\big[{^4\text{C}}_0\text{a}^4+{^4\text{C}}_1\text{a}^3\text{b}+{^4\text{C}}_2\text{a}^2\text{b}^2+{^4\text{C}}_3\text{ab}^3+{^4\text{C}}_4\text{ab}^4\big]\\-\big[{^4\text{C}}_0\text{a}^4-{^4\text{C}}_1\text{a}^3\text{b}+{^4\text{C}}_2\text{a}^2\text{b}^2-{^4\text{C}}_3\text{ab}^3+{^4\text{C}}_4\text{b}^4$

$={^4\text{C}}_0\text{a}^4+{^4\text{C}}_1\text{a}^3\text{b}+{^4\text{C}}_2\text{a}^2\text{b}^2+{^4\text{C}}_3\text{ab}^3+{^4\text{C}}_4\text{ab}^4\\-{^4\text{C}}_0\text{a}^4+{^4\text{C}}_1\text{a}^3\text{b}-{^4\text{C}}_2\text{a}^2\text{b}^2+{^4\text{C}}_3\text{ab}^3+{^4\text{C}}_4\text{b}^4\big]$

$=2\big[{^4\text{C}}_1\text{a}^3\text{b}+{^4\text{C}}_3\text{ab}^3\big]$

$=2\big[4\text{a}^3\text{b}+4\text{ab}^3\big]$

$=8\big[\text{a}^3\text{b}+\text{ab}^3\big]$

$\therefore(\text{a+b})^4-(\text{a-b})^4=8(\text{a}^3\text{b}+\text{ab}^3)$

Putting $\text{a}=\sqrt3$ and $\text{b}=\sqrt2$ in equation (i) we get

$(\sqrt3+\sqrt2)^4-(\sqrt3-\sqrt2)^4=8\Big[(\sqrt3)^3\times\sqrt2+(\sqrt3)\times(\sqrt2)^2\Big]$

$=8\Big[3\sqrt6+2\sqrt6\Big]$

$=8\times5\sqrt6$

$=40\sqrt6$

$\therefore(\sqrt{3}+\sqrt2)^4-(\sqrt{3}-\sqrt2)^440\sqrt6.$