Without actually calculating the cubes,find t | vedclass

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(D) We have the expression: $(0.2)^{3} - (0.3)^{3} + (0.1)^{3}$.This can be rewritten as: $(0.2)^{3} + (-0.3)^{3} + (0.1)^{3}$.Let $a = 0.2$,$b = -0.3

Vedclass · Accepted Answer

$-0.018$

Vedclass · Answer

(D) We have the expression: $(0.2)^{3} - (0.3)^{3} + (0.1)^{3}$.This can be rewritten as: $(0.2)^{3} + (-0.3)^{3} + (0.1)^{3}$.Let $a = 0.2$,$b = -0.3$,and $c = 0.1$.Now,calculate the sum $a + b + c$:$a + b + c = 0.2 + (-0.3) + 0.1 = 0.3 - 0.3 = 0$.We know the algebraic identity: If $a + b + c = 0$,then $a^{3} + b^{3} + c^{3} = 3abc$.Substituting the values of $a$,$b$,and $c$:$a^{3} + b^{3} + c^{3} = 3(0.2)(-0.3)(0.1)$.Calculating the product:$3 	imes 0.2 = 0.6$$0.6 	imes (-0.3) = -0.18$$-0.18 	imes 0.1 = -0.018$.Therefore,$(0.2)^{3} - (0.3)^{3} + (0.1)^{3} = -0.018$.

Without actually calculating the cubes,find the value of:
$(0.2)^{3} - (0.3)^{3} + (0.1)^{3}$

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Without actually calculating the cubes,find the value of:$(0.2)^{3} - (0.3)^{3} + (0.1)^{3}$