Question

Find the fourth proportional to $(a + b)^2 \times (a^3 - b^3), (a^3 + b^3), (a^2 - b^2)$.

• A. $\frac{(a^2-ab+b^2)}{(a+b)}$
• B. $\frac{(a^2+ab+b^2)}{(a-b)}$
• C. $\frac{(a^2-ab+b^2)}{(a^2+ab+b^2)}$
• D. $\frac{(a^2+ab+b^2)}{(a^2-b^2)}$

Answer 1

Answer: (C)

$\frac{(a^2-ab+b^2)}{(a^2+ab+b^2)}$

Answer 2

To find the fourth proportional $X$ to $P = (a + b)^2 (a^3 - b^3)$, $Q = (a^3 + b^3)$, and $R = (a^2 - b^2)$, we use the relationship $\frac{P}{Q} = \frac{R}{X}$. Solving for $X$, we have $X = \frac{Q \times R}{P}$.

Substituting the given expressions: $X = \frac{(a^3 + b^3)(a^2 - b^2)}{(a + b)^2 (a^3 - b^3)}$

Now we factorize using algebraic identities:

• $a^3 + b^3 = (a+b)(a^2-ab+b^2)$
• $a^2 - b^2 = (a-b)(a+b)$
• $a^3 - b^3 = (a-b)(a^2+ab+b^2)$

Substituting these factorizations into the expression for $X$:

$X = \frac{[(a+b)(a^2-ab+b^2)] \times [(a-b)(a+b)]}{(a+b)^2 \times [(a-b)(a^2+ab+b^2)]}$

Simplifying the numerator: $(a+b)(a+b) = (a+b)^2$.

$X = \frac{(a+b)^2 (a-b) (a^2-ab+b^2)}{(a+b)^2 (a-b) (a^2+ab+b^2)}$

Canceling common terms $(a+b)^2$ and $(a-b)$ from numerator and denominator yields:

$X = \frac{(a^2-ab+b^2)}{(a^2+ab+b^2)}$