Without calculating the cubes, find the following value of 0.23-0.33+0.13.

# Without calculating the cubes, find the following value of ${\left(0.2\right)}^{3}-{\left(0.3\right)}^{3}+{\left(0.1\right)}^{3}$.

1. A
$-0.801$
2. B
$-0.018$
3. C
$-0.108$
4. D
$-0.18$

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### Solution:

The given expression is ${\left(0.2\right)}^{3}-{\left(0.3\right)}^{3}+{\left(0.1\right)}^{3}$.
Therefore, rewrite the given expression as,
${\left(0.2\right)}^{3}-{\left(0.3\right)}^{3}+{\left(0.1\right)}^{3}={\left(0.2\right)}^{3}+{\left(-0.3\right)}^{3}+{\left(0.1\right)}^{3}$
Using identity, ${a}^{3}+{b}^{3}+{c}^{3}=3\mathit{abc}+\left(a+b+c\right)\left({a}^{2}+{b}^{2}+{c}^{2}-\mathit{ab}-\mathit{bc}-\mathit{ca}\right)$ to solve the given expression
substituting value of a = 0.2, b = -0.3 and c = 0.1 in a+b+c we get, $a+b+c=\left(0.2\right)+\left(-0.3\right)+\left(0.1\right)=0$.
Hence, above formula reduces to ${a}^{3}+{b}^{3}+{c}^{3}=3\mathit{abc}$.
Therefore, we get,
${\left(0.2\right)}^{3}+{\left(-0.3\right)}^{3}+{\left(0.1\right)}^{3}=3\left(0.2\right)\left(-0.3\right)\left(0.1\right)$
${\left(0.2\right)}^{3}+{\left(-0.3\right)}^{3}+{\left(0.1\right)}^{3}=-0.018$
Therefore, option 2 is  correct.

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