Let c - a - ib 3-47 i- where a - b - c - Z -and i -1- Then the value of a - b - c isA- 52B- 49- 57D- 63

C+47i=(a^3 3ab^2)+(3a^2b b^3)i ⇒ c=a^3 3ab^2 and 47=3a^2b b^3 47=b(3a^2 b^2) Since a,b∈ Z^+, ∴ b is a divisor of 47. ⇒ b=1 or b=47 nbsp; If b=47 nbsp; 3a^ ...

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Byju's Answer

Standard XII

Mathematics

Inequality of 2 Complex Numbers

Let c = a + i...

Question

Let $c = (a + i b)^{3} - 47 i,$ where $a, b, c \in Z^{+}$ and $i = \sqrt{- 1}$ . Then the value of $a + b + c$ is

52

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49

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57

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63

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Solution

The correct option is C $57$
$c + 47 i = (a^{3} - 3 a b^{2}) + (3 a^{2} b - b^{3}) i$
$\Rightarrow c = a^{3} - 3 a b^{2}$ and $47 = 3 a^{2} b - b^{3}$

$47 = b (3 a^{2} - b^{2})$
Since $a, b \in Z^{+},$
$∴ b$ is a divisor of $47.$
$\Rightarrow b = 1$ or $b = 47$

If $b = 47$
$3 a^{2} = 1 + 47^{2}$
Since, $1 + 47^{2}$ is not divisible by $3, b$ cannot be $47.$
$∴ b = 1$

$a = 4 (∵ a \in Z^{+})$
$c = a^{3} - 3 a b^{2} = 52$
$∴ a + b + c = 57$

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Let c=(a+ib)3−47i, where a,b,c∈Z+ and i=√−1. Then the value of a+b+c is

The correct option is C 57 c+47i=(a3−3ab2)+(3a2b−b3)i ⇒c=a3−3ab2 and 47=3a2b−b3 47=b(3a2−b2) Since a,b∈Z+, ∴b is a divisor of 47. ⇒b=1 or b=47 If b=47 3a2=1+472 Since, 1+472 is not divisible by 3, b cannot be 47. ∴b=1 a=4 (∵a∈Z+) c=a3−3ab2=52 ∴a+b+c=57

Let $c = (a + i b)^{3} - 47 i,$ where $a, b, c \in Z^{+}$ and $i = \sqrt{- 1}$ . Then the value of $a + b + c$ is