if a and b are +ve integer such thatN=(a+ib)3 -107i is a +ve integer .Find the value of N.where i2 = -1

Ans:Hello Student,Please find answer to your question belowSince N is a +ve integer.Then its imaginary part must be zero.

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Grade 12th passAlgebra

if a and b are +ve integer such that
N=(a+ib)³-107i is a +ve integer .Find the value of N.
where i²= -1

saurabh

11 Years agoGrade 12th pass

1 Answer

Jitender Singh

11 Years ago

Ans:
Hello Student,
Please find answer to your question below

$N = (a+ib)^{3} - 107i$
$(x+y)^{3} = x^{3}+y^{3} + 3xy(x+y)$
$N = a^{3}+(ib)^{3}+3iab(a+ib) - 107i$
$N = a^{3}+i^{2}.i(b)^{3}+3ia^{2}b+3i^{2}ab^{2} - 107i$
$i^{2} = -1$
$N = a^{3}-i(b)^{3}+3ia^{2}b-3ab^{2} - 107i$
$N = (a^{3}-3ab^{2})+i(-b^{3}+3a^{2}b - 107)$
Since N is a +ve integer.
Then its imaginary part must be zero.
$-b^{3}+3a^{2}b-107 = 0$
$3a^{2}b-b^{3}=107$
$b(3a^{2}-b^{2})=107$
$N = a^{3}-3ab^{2}$
$N = a(a^{2}-3b^{2})$
$N = a(9a^{2}-3b^{2}-8a^{2})$
$N = a(3(3a^{2}-b^{2})-8a^{2})$
$N = a(3(\frac{107}{b})-8a^{2})$
$N = \frac{321a}{b}-8a^{3}$