Question

Find the value of
${(\sqrt{x^2-a^2}+x)}^5-{(\sqrt{x^2-a^2}-x)}^5$

• A. $2x(16x^4-20x^2{a}^2+5a^4)$
• B. $2x(16x^5-20x^4{a}^2+5a^4)$
• C. $2x(12x^4-16x^2{a}^2+5a^4)$
• D. $2x(8x^4+20x^2{a}^2+6a^4)$

Answer 1

The correct option is A $2x(16x^4-20x^2{a}^2+5a^4)$

We can write the given equation as $(x+\sqrt{x^2-a^2}{)}^5+(x+(-\sqrt{x^2-a^2}){)}^5$

We can see that in the expansion, odd powers of $\sqrt{x^2-a^2}$ in $1^{st}\&2^{nd}$ will cancel each other while the others add up.

We get

$(x+\sqrt{x^2-a^2}{)}^5-(\sqrt{x^2-a^2}-x{)}^5=2\left(5x\right(x^2-a^2{)}^2+10(x^2-a^2)x^3+x^5)$

$=2x\left(5\right(x^2-a^2{)}^2+10(x^2-a^2)x^2+x^4)$

$=2x(16x^4-20x^2{a}^2+5a^4)$