Evaluating 27a³ + 64b³ Using Algebraic Identities

We are given the equations:

• $3a + 4b = 2$
• $ab = \frac{1}{36}$

We need to find the value of $27a^3 + 64b^3$. Notice that this expression is the sum of cubes:

$27a^3 + 64b^3 = (3a)^3 + (4b)^3$

Applying the Sum of Cubes Formula

Recall the sum of cubes formula: $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$.

Let $x = 3a$ and $y = 4b$. Applying the formula:

$(3a)^3 + (4b)^3 = (3a + 4b)((3a)^2 - (3a)(4b) + (4b)^2)$

$(3a)^3 + (4b)^3 = (3a + 4b)(9a^2 - 12ab + 16b^2)$

Calculating Intermediate Terms

We know $3a + 4b = 2$ and $ab = \frac{1}{36}$. We need to find the value of $9a^2 + 16b^2$.

Square the first given equation:

$(3a + 4b)^2 = 2^2$

$9a^2 + 2(3a)(4b) + 16b^2 = 4$

$9a^2 + 24ab + 16b^2 = 4$

Substitute the value of $ab$:

$9a^2 + 24\left(\frac{1}{36}\right) + 16b^2 = 4$

$9a^2 + \frac{2}{3} + 16b^2 = 4$

Solve for $9a^2 + 16b^2$:

$9a^2 + 16b^2 = 4 - \frac{2}{3} = \frac{12}{3} - \frac{2}{3} = \frac{10}{3}$

Final Calculation

Now substitute the values back into the expanded sum of cubes formula:

$27a^3 + 64b^3 = (3a + 4b)(9a^2 + 16b^2 - 12ab)$

$27a^3 + 64b^3 = (2)\left(\frac{10}{3} - 12\left(\frac{1}{36}\right)\right)$

$27a^3 + 64b^3 = (2)\left(\frac{10}{3} - \frac{12}{36}\right)$

$27a^3 + 64b^3 = (2)\left(\frac{10}{3} - \frac{1}{3}\right)$

$27a^3 + 64b^3 = (2)\left(\frac{9}{3}\right)$

$27a^3 + 64b^3 = (2)(3)$

$27a^3 + 64b^3 = 6$