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BINOMIAL THEOREM — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSBINOMIAL THEOREM3 Marks
Question
Find the value of ${({a^2} + \sqrt {{a^2} - 1} )^4} + {({a^2} - \sqrt {{a^2} - 1} )^4}$
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Answer

Putting $a^2 = x$ and $\sqrt {{a^2} - 1} = y$ we have
${({a^2} + \sqrt {{a^2} - 1} )^4} + {({a^2} - \sqrt {{a^2} - 1} )^4}$$ = {(x + y)^4} + {(x - y)^4}$
$ = {[^4}{C_0}{x^4}{ + ^4}{C_1}{x^3}y{ + ^4}{C_2}{x^2}{y^2}$${ + ^4}{C_3}x{y^3}{ + ^4}{C_4}{y^4}] + {[^4}{C_4}{x^4}{ - ^4}{C_1}{x^3}y$${ + ^4}{C_2}{x^2}{y^2}{ - ^4}{C_3}x{y^3}{ + ^4}{C_4}{y^4}]$
$ = 2{[^4}{C_0}{x^4}{ + ^4}{C_2}{x^2}{y^2}{ + ^4}{C_4}{y^4}]$$ = 2[{x^4} + 6{x^2}{y^2} + {y^4}]$
$ = 2[{({a^2})^4} + 6{({a^2})^2}{(\sqrt {{a^2} - 1} )^2}$$ + {(\sqrt {{a^2} - 1} )^4}]$
$ = 2[{a^8} + 6{a^4}({a^2} - 1) + {({a^2} - 1)^2}]$
$ = 2[{a^8} + 6{a^6} - 6{a^4} + {a^4} - 2{a^2} + 1]$
$ = 2[{a^8} + 6{a^6} - 5{a^4} - 2{a^2} + 1]$

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