Q

# Find (x + 1)^6 + (x - 1)^6 . Hence or otherwise evaluate (root of 2 + 1)^6 + ( root of 2 - 1) ^ 6 .

Q12.    Find $(x+1)^6 + (x-1)^6$ . Hence or otherwise evaluate $(\sqrt2+1)^6 + (\sqrt2-1)^6$.

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Using Binomial Theorem, the expressions $(x+1)^4$ and $(x-1)^4$ can be expressed as ,

$(x+1)^6=^6C_0x^6+^6C_1x^51+^6C_2x^41^2+^4C_3x^31^3+^6C_4x^21^4+^6C_5x1^5+^6C_61^6$

$(x-1)^6=^6C_0x^6-^6C_1x^51+^6C_2x^41^2-^4C_3x^31^3+^6C_4x^21^4-^6C_5x1^5+^6C_61^6$

From Here,

$\\(x+1)^6-(x-1)^6=^6C_0x^6+^6C_1x^51+^6C_2x^41^2+^4C_3x^31^3+$$^6C_4x^21^4+^6C_5x1^5+^6C_61^6$$\:\:\:\:\;\:\:\;\:\:\:\ +^6C_0x^6-^6C_1x^51+^6C_2x^41^2-^4C_3x^31^3+^6C_4x^21^4-^6C_5x1^5+^6C_61^6$

$(x+1)^6+(x-1)^6=2(^6C_0x^6+^6C_2x^41^2+^6C_4x^21^4+^6C_61^6)$

$(x+1)^6+(x-1)^6=2(x^6+15x^4+15x^2+1)$

Now, Using this, we get

$(\sqrt2+1)^6 + (\sqrt2-1)^6=2((\sqrt{2})^6+15(\sqrt{2})^4+15(\sqrt{2})^2+1)$

$(\sqrt2+1)^6 + (\sqrt2-1)^6=2(8+60+30+1)=2(99)=198$

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