Use Your Identities!

$x^3+y^3=(x+y)(x^2+y^2-xy)$
$x^3+y^3=(x+y)[(x+y)^2-3xy]$
$x^3+y^3=5×4=\boxed{20}$

Okay, so we have this identity: $(x+y)^{3}=x^{3}+3x^{2}y+3xy^{2}+y^{3}$ Here, we replace $(x+y)^{3}$ with 125 because $x+y=5$ . So, $125=x^{3}+3x^{2}y+3xy^{2}+y^{3}$ $125=x^{3}+y^{3}+3x^{2}y+3xy^{2}$ Note: $3x^{2}y+3xy^{2}=3xy(x+y)$ and that implies $3 \times 7 \times (5)=105$ $125=x^{3}+y^{3}+105$ $125-105=x^{3}+y^{3}$ Therefore, $x^{3}+y^{3}=\boxed{20}$

We know that; $(\large\color{#D61F06}x+\color{#3D99F6}y)^{3}=\color{#D61F06}x^{3}+\color{#3D99F6}y^{3}+3\color{#D61F06}x\color{#3D99F6}y(\color{#D61F06}x+\color{#3D99F6}y)$ $\large\therefore 5^{3}=\color{#D61F06}x^{3}+\color{#3D99F6}y^{3}+3\times7\times 5\\\large\color{#D61F06}x^{3}+\color{#3D99F6}y^{3}=125-105\\\large\color{#D61F06}x^{3}+\color{#3D99F6}y^{3}=\boxed{20}$

Hence the answer is $20$ .

Some very elegant solutions by working out (x+y)^3, but don't forget you could also show x = 7/y and x = 5-y so 7/y = 5-y or y^2-5y+7=0, solve for y, then solve for x. Harder solution, much less elegant, but perfectly legit (especially if you note that y!=0)

Using the identity;

$(x+y)^{3}=x^{3}+y^{3}+3xy(x+y)$

$5^{3}=x^{3}+y^{3}+3.7.(5)$

$x^{3}+y^{3}=125-105$

$x^{3}+y^{3}=20$

Thus, the answer is $20$ .

a^3 + b^3 = (a + b)^3 - 3 a b (a + b) ...enter the values.....5^3-3(5)(7)=125-105=20

xy=7 x+y=5

Then xy(x+y)=35

x^2y+xy^2=35

and (x+y)^3=125

x^3+3(x^2y+xy^2)+y^3=125

x^3+3(35)+y^3=125

x^3+y^3+105=125

x^3+y^3=20