Without calculating the cubes, find the value of -(0.4)3-(0.2)3+(0.6)3.

# Without calculating the cubes, find the value of $-{\left(0.4\right)}^{3}-{\left(0.2\right)}^{3}+{\left(0.6\right)}^{3}$.

1. A
0.144
2. B
0.4441
3. C
0.441
4. D
0.414

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### Solution:

It is given $-{\left(0.4\right)}^{3}-{\left(0.2\right)}^{3}+{\left(0.6\right)}^{3}$.
We know the formula,
${x}^{3}+{y}^{3}+{z}^{3}-3\mathit{xyz}=\left(x+y+z\right)\left({x}^{2}+{y}^{2}+{z}^{2}-\mathit{xy}-\mathit{yz}-\mathit{zx}\right)$ and
if then ${x}^{3}+{y}^{3}+{z}^{3}=3\mathit{xyz}$.
Rewriting the given expression as x = -0.4, y = -0.2, and z = 0.6,
=$-{\left(0.4\right)}^{3}-{\left(0.2\right)}^{3}+{\left(0.6\right)}^{3}$
$=3\left(-0.4\right)\left(-0.2\right)\left(0.6\right)$
$=3\left(-0.4\right)\left(-0.12\right)$
$=3\left(0.048\right)$
$=0.144$
Hence, the value of $-{\left(0.4\right)}^{3}-{\left(0.2\right)}^{3}+{\left(0.6\right)}^{3}=0.144.$
Therefore, option 1 is  correct.

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