If A = [ * 20 c 1 + a^2 + a^4 & 1 + ab + a^2 b^2 & 1 + ac + a^2 c… - Vidyadip

MCQ

If $A = \left[ {\begin{array}{*{20}{c}}
{1 + {a^2} + {a^4}}&{1 + ab + {a^2}{b^2}}&{1 + ac + {a^2}{c^2}} \\
{1 + ab + {a^2}{b^2}}&{1 + {b^2} + {b^4}}&{1 + bc + {b^2}{c^2}} \\
{1 + ac + {a^2}{c^2}}&{1 + bc + {b^2}{c^2}}&{1 + {c^2} + {c^4}}
\end{array}} \right]$

and $det(A) = det(4I)$, where $I$ is $3 × 3$ identity matrix, then $(a -b)^3 + (b -c)^3 + (c -a)^3$ can be equal to -

✓
$-24$
B
$6$
C
$-6$
D
$12$

✓

Answer

Correct option: A.

$-24$

a
$A=\left[\begin{array}{lll}{1} & {a} & {a^{2}} \\ {1} & {b} & {b^{2}} \\ {1} & {c} & {c^{2}}\end{array}\right]\left[\begin{array}{lll}{1} & {1} & {1} \\ {a} & {b} & {c} \\ {a^{2}} & {b^{2}} & {c^{2}}\end{array}\right]$

$\Rightarrow \quad \operatorname{det}(\mathrm{A})=(\mathrm{a}-\mathrm{b})^{2}(\mathrm{b}-\mathrm{c})^{2}(\mathrm{c}-\mathrm{a})^{2}$

and det $(4 \mathrm{I})=64$

$\Rightarrow \quad(a-b)(b-c)(c-a)=\pm 8$

$\because \quad(a-b)+(b-c)+(c-a)=0$

$\therefore \quad(a-b)^{3}+(b-c)^{3}+(c-a)^{3}$

$=3(a-b)(b-c)(c-a)=\pm 24$

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