If $$x + y = 4$$ and $$x^3 + y^3 = 12,$$ then the value of $$x^4 + y^4 = $$?

Given,  $$x^3 + y^3 = 12$$ ...........(1)

$$x + y = 4$$

$$=$$>  $$\left(x+y\right)^3=4^3$$

$$=$$>  $$x^3+y^3+3xy\left(x+y\right)=64$$

$$=$$>  $$12+3xy\left(4\right)=64$$

$$=$$>  $$12xy=52$$

$$=$$>  $$xy=\frac{52}{12}$$

$$=$$>  $$xy=\frac{13}{3}$$ ..........................(2)

$$x + y = 4$$

$$=$$>  $$\left(x+y\right)^2=4^2$$

$$=$$>  $$x^2+y^2+2xy=16$$

$$=$$>  $$x^2+y^2+2\left(\frac{13}{3}\right)=16$$

$$=$$>  $$x^2+y^2=16-\frac{26}{3}$$

$$=$$>  $$x^2+y^2=\frac{22}{3}$$..................(3)

$$\therefore\ $$ $$\left(x^3+y^3\right)\left(x+y\right)=\left(12\right)\left(4\right)$$

$$=$$>  $$x^4+x^3y+y^3x+y^4=48$$

$$=$$>  $$x^4+y^4+xy\left(x^2+y^2\right)=48$$

$$=$$>  $$x^4+y^4+\left(\frac{13}{3}\right)\left(\frac{22}{3}\right)=48$$

$$=$$>  $$x^4+y^4+\frac{286}{9}=48$$

$$=$$>  $$x^4+y^4=48-\frac{286}{9}$$

$$=$$>  $$x^4+y^4=\frac{432-286}{9}$$

$$=$$>  $$x^4+y^4=\frac{146}{9}$$

Hence, the correct answer is Option B