Firstly, the expression is simplified by using Binomial Theorem
\( (x+y)^4+(x-y)^4\)
This can be done as
= \( (x+y)^4=^4C_0x^4+\;^4C_1x^3y+\;\)
\( ^4C_2x^2y^2+\;^4C_3xy^3+\;^4C_4y^4\)
= \( x^4+4x^3y+6x^2y^2+4xy^3+y^4\)
= \( (x-y)^4=\;^4C_0x^4-\;^4C_1x^3y+\;\)
\( ^4C_2x^2y^2-\;^4C_3xy^3+\;^4C_4y^4\)
= \( x^4-4x^3y+6x^2y^2-4xy^3+y^4\)
Putting \( x=a^2\) and \( y=\sqrt{a^2-1}\), we obtain
= \( (a^2+\sqrt{a^2-1})^4+(a^2-\sqrt{a^2-1})^4\)
= \( 2[(a^2)^4+6(a^2)^2(\sqrt{a^2-1}^2+(\sqrt{a^2-1})^4]\)
= \( 2[a^8+6a^4(a^2-1)+(a^2-1)^2]\)
= \( 2[a^8+6a^4-6a^4+a^4-2a^2+1]\)
\( = 2[a^8+6a^6-5a^4-2a^2+1]\)
= \( 2a^8+12a^6-10a^4-4a^2+2\)
Answered by Abhisek | 1 year agoFind the term independent of x in the expansion of \( (\dfrac{3}{2x^2} – \dfrac{1}{3x})^9\)
Find the middle term in the expansion of \((x-\dfrac{ 1}{x})^{2n+1}\)
Find the middle term in the expansion of (1 + 3x + 3x2 + x3)2n
Find the middle term in the expansion of \( (\dfrac{x}{a} – \dfrac{a}{x})^{10}\)