If $$x^3+y^3=16$$ and $$x+y=4$$, then the value of $$x^4+y^4$$ is:

Given,  $$x^3+y^3=16$$ and $$x+y=4$$

$$\Rightarrow$$  $$\left(x+y\right)\left(x^2-xy+y^2\right)=16$$

$$\Rightarrow$$  $$\left(4\right)\left(x^2+2xy+y^2-3xy\right)=16$$

$$\Rightarrow$$  $$\left(x+y\right)^2-3xy=4$$

$$\Rightarrow$$  $$\left(4\right)^2-3xy=4$$

$$\Rightarrow$$  $$3xy=16-4$$

$$\Rightarrow$$  $$3xy=12$$

$$\Rightarrow$$  $$xy=4$$

$$\therefore\ $$ $$x^4+y^4=x^4+y^4+2x^2y^2-2x^2y^2$$

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

$$=\left[x^2+y^2+2xy-2xy\right]^2-2\left(4\right)^2$$

$$=\left[\left(x+y\right)^2-2xy\right]^2-32$$

$$=\left[\left(4\right)^2-2\left(4\right)\right]^2-32$$

$$=\left[16-8\right]^2-32$$

$$=64-32$$

$$=32$$

Hence, the correct answer is Option B