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 $x+y=12$

Step 1: Find the product $xy$

Using the algebraic identity for the sum of cubes:

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

Substituting the given values:

$468=(12)^{3}-3 xy(12)$

$468=1728-36 x y$

$36 x y=1728-468$

$36 xy=1260$

$x y=35$

Step 2: Find $x^{2}+y^{2}$

Using the identity for $(x+y)^{2}$ :

$(x+y)^{2}=x^{2}+2 x y+y^{2}$

$12^{2}=x^{2}+2(35)+y^{2}$

$144=x^{2}+70+y^{2}$

$x^{2}+y^{2}=74$

Step 3: Find $x^{4}+y^{4}$

Using the identity:

$x^{4}+y^{4}=\left(x^{2}+y^{2}\right)^{2}-2(x y)^{2}$

Substituting the values:

$x^{4}+y^{4}=(74)^{2}-2(35)^{2}$

$x^{4}+y^{4}=5476-2(1225)$

$x^{4}+y^{4}=5476-2450$

$x^{4}+y^{4}=3026$

Therefore, the value of $x^{4}+y^{4}=3026$ and the correct option is Option 4.