If (x+iy)^{frac{1}{3}} = 5+3i,then 3x+5y = | vedclass

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(D) Given $(x+iy)^{\frac{1}{3}} = 5+3i$. Cubing both sides,we get $x+iy = (5+3i)^3$. Using the identity $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$,we have:

Vedclass · Accepted Answer

$960$

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(D) Given $(x+iy)^{\frac{1}{3}} = 5+3i$. Cubing both sides,we get $x+iy = (5+3i)^3$. Using the identity $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$,we have: $x+iy = 5^3 + 3(5^2)(3i) + 3(5)(3i)^2 + (3i)^3$. $x+iy = 125 + 3(25)(3i) + 15(9i^2) + 27i^3$. Since $i^2 = -1$ and $i^3 = -i$,we get: $x+iy = 125 + 225i + 135(-1) + 27(-i)$. $x+iy = 125 + 225i - 135 - 27i$. $x+iy = (125-135) + (225-27)i$. $x+iy = -10 + 198i$. Comparing real and imaginary parts,$x = -10$ and $y = 198$. Now,calculate $3x+5y = 3(-10) + 5(198) = -30 + 990 = 960$.

If $(x+iy)^{\frac{1}{3}} = 5+3i$,then $3x+5y = $

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