Without actually calculating the cubes,find t | vedclass

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(C) Let $a = 0.2$,$b = -0.3$,and $c = 0.1$.We observe that $a + b + c = 0.2 + (-0.3) + 0.1 = 0.3 - 0.3 = 0$.We know the algebraic identity: If $a + b

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$-0.018$

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(C) Let $a = 0.2$,$b = -0.3$,and $c = 0.1$.We observe that $a + b + c = 0.2 + (-0.3) + 0.1 = 0.3 - 0.3 = 0$.We know the algebraic identity: If $a + b + c = 0$,then $a^{3} + b^{3} + c^{3} = 3abc$.Substituting the values:$(0.2)^{3} + (-0.3)^{3} + (0.1)^{3} = 3(0.2)(-0.3)(0.1)$.Calculating the product: $3 	imes 0.2 = 0.6$; $0.6 	imes (-0.3) = -0.18$; $-0.18 	imes 0.1 = -0.018$.Thus,$(0.2)^{3} - (0.3)^{3} + (0.1)^{3} = -0.018$.

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

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