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 "-" , all odd powers of will cancel. And since the powers must total 6, that means all odd powers of will also. That's why the answer is an integer. Of course, the even powers will be doubled.
So you will have .
$(a+b)^6=a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^ 5+b^6$
$(a-b)^6=a^6-6a^5b+15a^4b^2-20a^3b^3+15a^2b^4-6ab^5+b^6$
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.