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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 , 10 Years ago
Grade 12th pass
anser 1 Answers
Jitender Singh

Last Activity: 10 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}


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