Question

If x=2+3i and y=2-3i then the value of x^3+y^3 is​

Answer 1

Consider the given expressions.

3+i(1+i)x−2i​+3−i(2−3i)y+i​=i

3+ix+(x−2)i​+3−i2y+(1−3y)i​=i

9−i2(x+(x−2)i)(3−i)+(2y+(1−3y)i)(3+i)​=i

x(3−i)+i(x−2)(3−i)+2y(3+i)+i(1−3y)(3+i)=(9+1)i

3x−ix+i(3x−ix−6+2i)+6y+2iy+i(3+i−9y−3yi)=10i

3x−ix+3xi−i2x−6i+2i2+6y+2iy+3i+i2−9yi−3yi2=10i

3x−ix+3xi+x−6i−2+6y+2iy+3i−1−9yi+3y=10i

4x+9y−3+2xi−7yi−13i=0

4x+9y−3+(2x−7y−13)i=0

On comparing real part and imaginary part, we get

4x+9y−3=0     …… (1)

2x−7y−13=0  …… (2)

On solving both equations, we get

x=3

y=−1

Hence, the value of x,y is 3,−1.

Step-by-step explanation: