How to derive $\;a^3 + b^3 = (a + b)(a^2 - ab + b^2)\;?$

Question

I'm studying in 8th grade in India. I saw the question in my FIITJEE textbook, but it was not shown how to derive it. Please help. My mid phase examinations are coming up.

Answer 1

$$a^3+b^3=(a+b)^3-3ab(a+b)=(a+b)\{(a+b)^2-3ab\}=\cdots$$

Answer 2

$(a+b)(a^2-ab+b^2)=a^3-a^2b+ab^2+a^2b-ab^2+b^3=a^3+b^3$

Answer 3

Start with $$\color{grey}{\boxed{\displaystyle\color{white}{\overline{\underline{\color{black}{\,a^n-c^n=\left(a-c\right)\left( a^{n-1}+a^{n-2}c+\cdots+ac^{n-2}+c^{n-1}\right)\quad n\in\mathbb N.\,}}}}}}$$ and set $n=3$ and $c=-b.$

Answer 4

In general (and straighforward):

$$a^{n}-b^{n}=\left(a-b\right)\left(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}\right)$$

If $n$ is odd then $\left(-b\right)^{n}=-b^{n}$ and application gives: $$a^{n}+b^{n}=a^{n}-\left(-b\right)^{n}=(a+b)(\cdots)$$

Answer 5

Perhaps you were confused or misled by the word "derive" here. Sometimes, when asked to derive or prove an equation, all that is being asked for is that you check that the two sides of the equation are indeed equal by 'multiplying out' any complex terms (the right hand side in this case) and ending up with the same term on each side. So @JasperLoy's simple answer would (by my lights) do just fine.

Answer 6

Note that $-1$ is a root of polynomial $f = x^{3} + 1$.

Thus $x+ 1$ divides $f$.

Do long division to find $$ x^{3} + 1 = (x+1)(x^{2} - x + 1). $$ Now substitute $x = a/b$ and multiply both sides by $b^{3}$ to get the equality.

Of course the equality is trivial if $b = 0$.

Answer 7

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

You just multiply it and get:

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

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

So:

$$a^3+b^3=(a+b)^3-(3a^2 b+3ab^2)=(a+b)^3-3[ab(a+b)]$$

You factor out $(a+b)$:

$$(a+b)^3-3(ab(a+b))= (a+b)^3-(a+b)3ab=(a+b)[(a+b)^2-3ab]$$

When you expand (a+b) squared you get

$$(a+b)[(a+b)^2-3ab]=(a+b)(a^2-ab+b^2).$$