Question

If \(x, y\) are two real numbers such that \(x+y=23\) and \(x^{2}+y^{2}=289\), then \(x^{3}+y^{3}=\)

• A. 529
• B. 2817
• C. 6647
• D. 3887

Hint:

First find \(xy\) from \[ (x+y)^2=x^2+y^2+2xy \] and then use \[ x^3+y^3=(x+y)(x^2-xy+y^2). \]

Answer 1

The Correct Option is D

Step 1: Find the value of \(xy\).
Using the identity \[ (x+y)^2=x^2+y^2+2xy \] Substituting the given values, \[ 23^2=289+2xy \] \[ 529=289+2xy \] \[ 2xy=240 \] \[ xy=120 \]

Step 2: Apply the identity for the sum of cubes.
We know that \[ x^3+y^3=(x+y)(x^2-xy+y^2) \] Substituting the values, \[ x^3+y^3=23(289-120) \] \[ =23(169) \]

Step 3: Perform the multiplication.
\[ 23\times169 = 23(100+60+9) \] \[ =2300+1380+207 \] \[ =3887 \] {3887}