how to solve question no.12 without using calculator?...my final calculation was 64(21∧3/2)..but i got it wrong as the answer given is 7040
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First, because of the "+" and "-" $\displaystyle \sqrt{3}$, all odd powers of $\displaystyle \sqrt{3}$ will cancel. And since the powers must total 6, that means all odd powers of $\displaystyle \sqrt{7}$ will also. That's why the answer is an integer. Of course, the even powers will be doubled.
So you will have $\displaystyle 2\left((\sqrt{7})^6+ \frac{6!}{2!4!}(\sqrt{7})^4(\sqrt{3})^2+ \frac{6!}{4!2!}(\sqrt{7})^2(\sqrt{3})^4+ (\sqrt{3})^6\right)= 2\left(343+ 15(49)(3)+ 15(7)(9)+ 27\right)= 2(343+ 2205+ 945+ 27)= 2(3520)= 7040$.
$(a+b)^6=a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^ 5+b^6$
$(a-b)^6=a^6-6a^5b+15a^4b^2-20a^3b^3+15a^2b^4-6ab^5+b^6$
sum ...
$2a^6+30a^4b^2+30a^2b^4+2b^6$
$2[a^6+15a^2b^2(a^2+b^2)+b^6]$
sub in $\sqrt{7}$ for $a$ and $\sqrt{3}$ for $b$ ...
$2[7^3+315(10)+3^3]$
$2[343+3150+27]$
$2[370+3150]$
$2[3520]=7040$
... contrary to what you may think, this problem is not "advanced algebra" ... it's regular algebra or precalculus at most.