Question

Find (x + 1)^6 + (x - 1)^6. Hence or otherwise evaluate (√2+ 1)^6 +(√2- 1)^6.
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Answer 1

MARK ME AS THE BRAINLIEST

Answer:

x^12 + 1

Step-by-step explanation:

(x + 1)^6 + (x - 1)^6 = (x + 1)^6+6

= (x+1)^12

= x^12 + 1

HOPE IT IS USEFUL

Answer 2

$\large\text{\underline{Footnote}}$

This is easily solvable with substitution.

$\large\text{\underline{Solution}}$

First, we are going to use the substitution method.

Substitute $A$ and $B$ for $(x+1)^{2}$ and $(x-1)^{2}$ respectively.

By factorization,

$\implies A^{3}+B^{3}=(A+B)(A^{2}-AB+B^{2})$

Now, we usually find the two values, the sum, and the product;

$\implies A+B=2x^{2}+2$

$\implies AB=(x^{2}-1)^{2}=x^{4}-2x^{2}+1$

However,

$\implies A^{2}-AB+B^{2}=(A+B)^{2}-3AB$

$\implies A^{2}-AB+B^{2}=(2x^{2}+2)^{2}-3(x^{4}-2x^{2}+1)$

$\implies A^{2}-AB+B^{2}=(4x^{4}+8x^{2}+4)+(-3x^{4}+6x^{2}-3)$

$\implies A^{2}-AB+B^{2}=x^{4}+14x^{2}+1$

Hence,

$\implies A^{3}+B^{3}=(2x^{2}+2)(x^{4}+14x^{2}+1)$

$\implies(x+1)^{6}+(x-1)^{6}=(2x^{2}+2)(x^{4}+14x^{2}+1)$

To solve the question we substitute $x=\sqrt{2}$, so we are substituting the value of $x^{2}$, which is 2.

$\implies(\sqrt{2}+1)^{6}+(\sqrt{2}-1)^{6}=(4+2)\times(4+28+1)$

$\implies(\sqrt{2}+1)^{6}+(\sqrt{2}-1)^{6}=6\times33$

$\implies(\sqrt{2}+1)^{6}+(\sqrt{2}-1)^{6}=198$

$\large\text{\underline{Conclusion}}$

The required value is 198.