JIPMAT 2025 (QA) - If x^3 + y^3 = 468 and x + y = 12, then value of x^4 + y^4 will be | PYQs + Solutions

JIPMAT 2025

Algebra

Identities

Medium

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

3620

2036

3025

3026

✅ Correct Option: 4

Given:

\newline

x^{3}+y^{3}=468

and

x+y=12

Step 1: Find the product

xy

\newline

Using the algebraic identity for the sum of cubes:

\newline

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

Substituting the given values:

\newline

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

\newline

468=1728-36 x y

\newline

36 x y=1728-468

\newline

36 xy=1260

\newline

x y=35

Step 2: Find

x^{2}+y^{2}

\newline

Using the identity for

(x+y)^{2}

\newline

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

\newline

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

\newline

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

\newline

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

Step 3: Find

x^{4}+y^{4}

\newline

Using the identity:

\newline

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

Substituting the values:

\newline

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

\newline

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

\newline

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

\newline

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

Therefore, the value of

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

and the correct option is Option 4.

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