Question

The value of \[ \frac{15^3 + 6^3 + 3 \cdot 6 \cdot 15 \cdot 21}{1 + 4(6) + 6(36) + 4(216) + 1296} \] is equal to: 

• A. \( \frac{29}{7} \)
• B. \( \frac{7}{19} \)
• C. \( \frac{6}{17} \)
• D. \( \frac{21}{19} \)
• E. \( \frac{27}{7} \)

Hint:

Recognizing standard identities can save a lot of time in lengthy algebraic expressions.

Answer 1

The Correct Option is D

Concept: Use identities: \[ a^3 + b^3 + 3ab(a+b) = (a+b)^3 \] and \[ 1 + 4x + 6x^2 + 4x^3 + x^4 = (1+x)^4 \]

Step 1: Simplify the numerator.
\[ 15^3 + 6^3 + 3 \cdot 6 \cdot 15 \cdot 21 \] Since \( 21 = 15 + 6 \), use identity: \[ a^3 + b^3 + 3ab(a+b) = (a+b)^3 \] \[ = (15 + 6)^3 = 21^3 \]

Step 2: Simplify the denominator.
\[ 1 + 4(6) + 6(36) + 4(216) + 1296 \] \[ = 1 + 4x + 6x^2 + 4x^3 + x^4 \quad \text{where } x=6 \] \[ = (1+6)^4 = 7^4 \]

Step 3: Combine the result.
\[ \frac{21^3}{7^4} \] \[ = \frac{(3 \cdot 7)^3}{7^4} = \frac{27 \cdot 7^3}{7^4} \]

Step 4: Simplify powers.
\[ = \frac{27}{7} \]

Step 5: Final answer.
\[ \boxed{\frac{27}{7}} \]