Would You Like a Cube With That?

$a^3 + b^3 = 2593080$ $\color{#3D99F6}{a + b = 210}$ $a^3+b^3=2593080$

$(\color{#3D99F6}{a+b})(a^2-ab+b^2)=2593080$

$(\color{#3D99F6}{210})(a^2-ab+b^2)=2593080$

$a^2-ab+b^2=12348$

$a^2-ab\color{#D61F06}{+3ab}+b^2=12348\color{#D61F06}{+3ab}$

$a^2+2ab+b^2=12348+3ab$

$(\color{#3D99F6}{a+b})^2=12348+3ab$

$\color{#3D99F6}{210}^2=12348+3ab$

$44100=12348+3ab$

$31752=3ab$

$\boxed{ab=10584}$

There is another way to solve this, it is as follows.

we know that, ${ (a+b) }^{ 3 }={ a }^{ 3 }+{ b }^{ 3 }+3ab(a+b)$

Substituting the values in the equation above we get, ${ (210) }^{ 3 }=2593080+3ab(210)$

Or, $9261000-2593080=3ab(210)$

Or, $6667920=3ab(210)$

Or, $\displaystyle 3ab=\frac { 6667920 }{ 210 } =31752$

Or, $\displaystyle ab=\frac { 31752 }{ 3 }$

Or, $ab= \boxed{10584}$

$\begin{aligned} a^3+b^3 & = (a+b)(a^2+b^2-ab) = (a+b)[(a+b)^2-2ab-ab] \\ & = (a+b)[(a+b)^2-3ab] \\ \Rightarrow 2593080 & = 210(210^2 - 3ab) \\ \Rightarrow ab & = \dfrac {1}{3} \left( 210^2 - \dfrac {2593080}{210} \right) = \dfrac {44100-12348}{3} = \dfrac {31752}{3} \\ & = \boxed{10584} \end{aligned}$

the solution is very simple based on the factorization of ${a}^3+{b}^3$

so as we know that $a^3+b^3=(a+b)(a^2-ab+b^2)$ , keep in your mind we

need $a+b$ and not $a-b$ so we will add $-2ab+2ab=0$ to the second term to get :

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

by using ${(a+b)}^{2}=a^2+b^2+2ab$

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

back word substitution to get :

$2593080=(210)((210)^2-3ab)$

thus $12348=44100-3ab$

therefore $ab=\frac{12384-44100}{-3}=10584$

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

$2593080 =(a+b)(a^2+2ab+b^2)$

$2593080= (120)(a^2+2ab+b^2)$

$12348= (a^2+2ab+b^2) --- (1)$

We know that

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

$a^2+2ab+b^2 = 210*210 = 44100 --- (2)$

Use systems of linear Equation on (1) and (2)

$3ab = 31752$

$ab= 10584$

a^2 + b^2 + 2 a b = 210^2 ................... (1)

a^2 + b^2- a b = 2593080/210 ..........(2)

(1) - (2)

3 a b = 31752

a b = 10584

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

$2593080=210(a^2-ab+b^2)$

$\dfrac{2593080}{210}=a^2-ab+b^2$

My own derived identity: $(a+b)^2 - 3ab=$ $\color{#3D99F6}210^2-3ab$ $=a^2+2ab+b^2-3ab=$ $\color{#3D99F6}a^2-ab+b^2$

Substitute:

$\dfrac{2593080}{210}=210^2-3ab$

$3ab=31752$

$\large \color{#D61F06}\boxed{ab=10584}$

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

$2593080 = (210)(a^{2} - ab + b^{2})$

$12348 = a^{2} - ab + b^{2}$

$ab = ( a^{2} + b^{2} ) - 12348$

In order to solve the problem, you'll need to get the value of $a^{2} + b^{2}$ . We know that,

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

$44100 = a^{2} + 2ab + b^{2}$

$44100 - 2ab = a^{2} + b^{2}$

Now substitute the value of $a^{2} + b^{2}$ to the equation.

$ab = (44100 - 2ab) - 12348$

$3ab = 31752$

$ab = \boxed{10584}$

Dk why the number are so big (needed to use a calculator (T-T))

A = 126 and b is 84.