Question

Derive the value of a3+b3 from ( a+b)3 and what is the relation between (a+b)2 and( a-b) means find the solution of (a+b)2 in form( a-b)

Answer 1

First of all let us know what is $(a+b)^3$

$(a+b)^3= (a+b) (a+b) (a+b)$

$={(a+b) (a+b)} (a+b)$

$={a(a+b) + b(a+b)} (a+b)$

$=(a^2 + ab + ab + b^2) (a+b)$

$=(a^2 + b^2 + 2ab) (a+b)$

$=a^2(a+b) + b^2(a+b) + 2ab(a+b)$

$=a^3 + a^2b + ab^2 + b^3 + 2a^2b + 2ab^2$

$=a^3 + b^3 + 3a^2b + 3ab^2$

$=a^3 + b^3 + 3ab(a+b)$

Now when we have expanded  $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$

We can equate it

$(a+b)^3 = a^3 + b^3 + 3ab(a+b)$

$(a+b)^3 - 3ab(a+b) = a^3 + b^3$

$\therefore a^3 + b^3 = (a+b)^3 - 3ab(a+b)$

The following is the relation between $(a+b)^2$ and $( a-b)$

$(a+b)^2=(a-b)+a(a-1)+b(2a+b+1)$

 The solution of $(a+b)2$ in form $( a-b)$

$(a+b)^2=(a+b)(a+b)$

$=a(a+b)+b(a+b)$

$=a^2+ab+ab+b^2+a-a+b-b$

Rearranging gives

$(a+b)^2=(a-b)+a^2-a+2ab+b^2+b$

Factor common terms

$\therefore (a+b)^2=(a-b)+a(a-1)+b(2a+b+1)$