Evaluate.
Question:

Evaluate $(\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}$

Solution:

Firstly, the expression $(a+b)^{6}-(a-b)^{6}$ is simplified by using Binomial Theorem.

This can be done as

$(a+b)^{6}={ }^{6} \mathrm{C}_{0} a^{6}+{ }^{6} \mathrm{C}_{1} a^{5} b+{ }^{6} \mathrm{C}_{2} a^{4} b^{2}+{ }^{6} \mathrm{C}_{3} a^{3} b^{3}+{ }^{6} \mathrm{C}_{4} a^{2} b^{4}+{ }^{6} \mathrm{C}_{5} a^{1} b^{5}+{ }^{6} \mathrm{C}_{6} b^{6}$

$=a^{6}+6 a^{5} b+15 a^{4} b^{2}+20 a^{3} b^{3}+15 a^{2} b^{4}+6 a b^{5}+b^{6}$

$(a-b)^{6}={ }^{6} \mathrm{C}_{6} a^{6}-{ }^{6} \mathrm{C}_{1} a^{5} b+{ }^{6} \mathrm{C}_{2} a^{4} b^{2}-{ }^{6} \mathrm{C}_{3} a^{3} b^{3}+{ }^{6} \mathrm{C}_{4} a^{2} b^{4}-{ }^{6} \mathrm{C}_{5} a^{1} b^{5}+{ }^{6} \mathrm{C}_{6} b^{6}$

$(a-b)^{6}={ }^{6} C_{0} a^{6}-{ }^{6} C_{1} a^{5} b+{ }^{6} C_{2} a^{4} b^{2}-{ }^{6} C_{3} a^{3} b^{3}+{ }^{6} C_{4} a^{2} b^{4}-{ }^{6} C_{5} a^{1} b^{5}+{ }^{6} C_{6} b^{6}$

$=a^{6}-6 a^{5} b+15 a^{4} b^{2}-20 a^{3} b^{3}+15 a^{2} b^{4}-6 a b^{5}+b^{6}$

$\therefore(a+b)^{6}-(a-b)^{6}=2\left[6 a^{5} b+20 a^{3} b^{3}+6 a b^{5}\right]$

Putting $a=\sqrt{3}$ and $b=\sqrt{2}$, we obtain

$(\sqrt{3}+\sqrt{2})^{6}-(\sqrt{3}-\sqrt{2})^{6}=2\left[6(\sqrt{3})^{5}(\sqrt{2})+20(\sqrt{3})^{3}(\sqrt{2})^{3}+6(\sqrt{3})(\sqrt{2})^{5}\right]$

$=2[54 \sqrt{6}+120 \sqrt{6}+24 \sqrt{6}]$

$=2 \times 198 \sqrt{6}$

$=396 \sqrt{6}$

 

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