If $$x^{3}+y^{3}=468$$ and x+y=12, then value of $$x^{4}+y^{4}$$ will be

Given, 

$$x^{3}+y^{3}=468$$ and x+y=12

We know,

$$x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)$$

Now, $$12\left(x^2+y^2-xy\right)=468$$

or, $$x^2+y^2-xy=39$$ -->(1)

or, $$\left(x+y\right)^2-2xy-xy=39$$

or, $$12^2-3xy=39$$

or, $$3xy=144-39=105$$

or, $$3xy=105$$

or, $$xy=35$$

Now, from equation (1),

$$x^2+y^2-35=39$$

or, $$x^2+y^2=\ 39+35\ =\ 74$$

So, $$x^4+y^4=\ \left(x^2+y^2\right)^2-2x^2y^2=74^2-2\left(35\right)^2$$

or, $$x^4+y^4=74^2-2\left(35\right)^2$$

or, $$x^4+y^4=5476-2450$$

or, $$x^4+y^4=3026$$