If (3+sqrt{2})^6-(3-sqrt{2})^6=a+b sqrt{2},th | vedclass

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(A) Using the binomial expansion $(x+y)^n - (x-y)^n = 2 \left[ \binom{n}{1} x^{n-1} y + \binom{n}{3} x^{n-3} y^3 + \binom{n}{5} x^{n-5} y^5 \right]$.H

Vedclass · Accepted Answer

$5220$

Vedclass · Answer

(A) Using the binomial expansion $(x+y)^n - (x-y)^n = 2 \left[ \binom{n}{1} x^{n-1} y + \binom{n}{3} x^{n-3} y^3 + \binom{n}{5} x^{n-5} y^5 \right]$.Here $x=3, y=\sqrt{2}, n=6$.$(3+\sqrt{2})^6 - (3-\sqrt{2})^6 = 2 \left[ \binom{6}{1} (3)^5 (\sqrt{2}) + \binom{6}{3} (3)^3 (\sqrt{2})^3 + \binom{6}{5} (3)^1 (\sqrt{2})^5 \right]$.$= 2 \left[ 6 \cdot 243 \cdot \sqrt{2} + 20 \cdot 27 \cdot 2\sqrt{2} + 6 \cdot 3 \cdot 4\sqrt{2} \right]$.$= 2 \left[ 1458\sqrt{2} + 1080\sqrt{2} + 72\sqrt{2} \right]$.$= 2 \left[ 2610\sqrt{2} \right] = 5220\sqrt{2}$.Comparing with $a+b\sqrt{2}$,we get $a=0$ and $b=5220$.Therefore,$a+b = 0 + 5220 = 5220$.

If $(3+\sqrt{2})^6-(3-\sqrt{2})^6=a+b \sqrt{2}$,then $a+b=$

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