If x^{a} cdot x^{b} cdot x^{c} = 1,then the v | vedclass

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(D) Given the equation $x^{a} \cdot x^{b} \cdot x^{c} = 1$.
Using the laws of exponents,$x^{a+b+c} = 1$.
Since $1 = x^{0}$,we have $x^{a+b+c} = x^{0

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$3abc$

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(D) Given the equation $x^{a} \cdot x^{b} \cdot x^{c} = 1$.
Using the laws of exponents,$x^{a+b+c} = 1$.
Since $1 = x^{0}$,we have $x^{a+b+c} = x^{0}$.
Equating the exponents,we get $a + b + c = 0$.
We know the algebraic identity: $a^{3} + b^{3} + c^{3} - 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)$.
Substituting $a + b + c = 0$ into the identity:
$a^{3} + b^{3} + c^{3} - 3abc = (0)(a^{2} + b^{2} + c^{2} - ab - bc - ca) = 0$.
Therefore,$a^{3} + b^{3} + c^{3} = 3abc$.

If $x^{a} \cdot x^{b} \cdot x^{c} = 1$,then the value of $a^{3} + b^{3} + c^{3}$ is

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