Let `z = x + iy` be a complex number where x and y are integers. Then ther area of the rectangle whose vertices are the roots of the equaiton `barz z^(3) + zbarz^(3) = 350`.

Explanation

We have,
`zbarz^(3)+barzz^(3)=350`
`rArr |z^(2)|barz^(2)+(barzz)z^(2)=350`
`rArr 2(x^(2)+y^(2))(x^(2)-y^(2))=350`, where `z=x+iy`
`rArr x^(2)=y^(2)=25` and `x^(2)-y^(2)=7`
`rArr x=+4` and `y=+-3`
Hence, the verticles of the rectangle are `A(-4,-3), B(4,-3), C(4,3)` and D(-4,3).
`therefore` Area of the rectangle ABCD = `AB xx CD = 8 xx 6`
=48 sq. units.