Question : What is the value of $\frac{0.74 \times 1.23 \times 0.13}{(0.37)^3+(0.41)^3-8(0.39)^3}$?
Option 1: $1$
Option 2: $-\frac{1}{3}$
Option 3: $\frac{1}{3}$
Option 4: $-1$
Correct Answer: $-\frac{1}{3}$
Solution :
$\frac{0.74 \times 1.23 \times 0.13}{(0.37)^3+(0.41)^3-8(0.39)^3}$
$=\frac{0.74 \times 1.23 \times 0.13}{(0.37)^3+(0.41)^3-(0.78)^3}$
We know the identity,
If $a+b+c=0$ then $a^3+b^3+c^3=3abc$
Here, $(0.37 + 0.41 - 0.78) = 0 ⇒ (0.37)^3+(0.41)^3-(0.78)^3=-3\times 0.37\times 0.41\times 0.78$
So, $\frac{0.74 \times 1.23 \times 0.13}{(0.37)^3+(0.41)^3-(0.78)^3}$
$=\frac{0.74 \times 1.23 \times 0.13}{-3\times 0.37\times 0.41\times 0.78}$
$=-\frac{1}{3}$
Hence, the correct answer is $-\frac{1}{3}$.
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