If a, b in mathbb{R} and i=sqrt{-1},then the | vedclass

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(D) Given the condition $(a+bi)^3 = a-bi$. Expanding the left side: $a^3 + 3a^2(bi) + 3a(bi)^2 + (bi)^3 = a-bi$. $a^3 + 3a^2bi - 3ab^2 - ib^3 = a-bi$.

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$5$

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(D) Given the condition $(a+bi)^3 = a-bi$. Expanding the left side: $a^3 + 3a^2(bi) + 3a(bi)^2 + (bi)^3 = a-bi$. $a^3 + 3a^2bi - 3ab^2 - ib^3 = a-bi$. Grouping real and imaginary parts: $(a^3 - 3ab^2) + i(3a^2b - b^3) = a - bi$. Equating real and imaginary parts: $1) a^3 - 3ab^2 = a \Rightarrow a(a^2 - 3b^2 - 1) = 0$. $2) 3a^2b - b^3 = -b \Rightarrow b(3a^2 - b^2 + 1) = 0$. Case $1$: If $a=0$,then $b(0 - b^2 + 1) = 0 \Rightarrow b(1-b^2) = 0$. So $b=0, 1, -1$. Pairs: $(0,0), (0,1), (0,-1)$. Case $2$: If $b=0$,then $a(a^2 - 1) = 0 \Rightarrow a=0, 1, -1$. Pairs: $(0,0), (1,0), (-1,0)$. Case $3$: If $a 
eq 0$ and $b 
eq 0$,then $a^2 - 3b^2 = 1$ and $3a^2 - b^2 = -1$. From the first,$a^2 = 1 + 3b^2$. Substitute into the second: $3(1 + 3b^2) - b^2 = -1$ $\Rightarrow 3 + 9b^2 - b^2 = -1$ $\Rightarrow 8b^2 = -4$,which has no real solution for $b$. Combining the solutions from Case $1$ and Case $2$,the distinct pairs are $(0,0), (0,1), (0,-1), (1,0), (-1,0)$. There are $5$ such ordered pairs.

If $a, b \in \mathbb{R}$ and $i=\sqrt{-1}$,then the number of ordered pairs of real numbers $(a, b)$ satisfying the condition $(a+bi)^3 = a-bi$ is

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