If (a3+ b3) is proportional to (a2 – b2), then (a2 - ab + b2) is proportional to
Understanding Proportionality in Algebra
The question deals with the concept of proportionality between algebraic expressions. When we say an expression A is proportional to an expression B, it means that A is equal to a constant multiple of B. Mathematically, this is written as \(A \propto B\), which is equivalent to \(A = k \cdot B\), where \(k\) is a constant of proportionality.
We are given that \((a^3 + b^3)\) is proportional to \((a^2 – b^2)\). Let's write this relationship using a constant \(k\):
\((a^3 + b^3) = k(a^2 – b^2)\)
Our goal is to find out what \((a^2 - ab + b^2)\) is proportional to. To do this, we can use the factorization formulas for the sum of cubes and the difference of squares.
Factorizing the Expressions
Recall the important algebraic factorization formulas:
- Sum of Cubes: \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\)
- Difference of Squares: \(a^2 - b^2 = (a-b)(a+b)\)
Now, let's substitute these factorized forms into our proportionality equation:
\((a+b)(a^2 - ab + b^2) = k(a-b)(a+b)\)
Solving for the Relationship
We have the equation \((a+b)(a^2 - ab + b^2) = k(a-b)(a+b)\). Assuming that \((a+b)\) is not equal to zero (if \(a+b=0\), the original expression \(a^2-b^2\) would be 0, leading to potential issues or trivial cases), we can divide both sides of the equation by \((a+b)\).
Dividing both sides by \((a+b)\), we get:
\(\frac{(a+b)(a^2 - ab + b^2)}{(a+b)} = \frac{k(a-b)(a+b)}{(a+b)}\)
This simplifies to:
\((a^2 - ab + b^2) = k(a-b)\)
Identifying the Proportionality
The resulting equation \((a^2 - ab + b^2) = k(a-b)\) shows that the expression \((a^2 - ab + b^2)\) is equal to a constant \(k\) multiplied by the expression \((a-b)\). By the definition of proportionality, this means that \((a^2 - ab + b^2)\) is proportional to \((a-b)\).
Let's look at the given options:
- (a - b)
- (a + b)
- (a + ab + b)
- (a³ – b³)
Comparing our result \((a^2 - ab + b^2) = k(a-b)\) with the options, we see that \((a^2 - ab + b^2)\) is proportional to \((a - b)\).
Thus, the correct option is (a - b).
| Expression | Factorization |
|---|---|
| \(a^3 + b^3\) | \((a+b)(a^2 - ab + b^2)\) |
| \(a^2 - b^2\) | \((a-b)(a+b)\) |
Revision Table: Key Algebra Concepts
| Concept | Definition/Formula | Example |
|---|---|---|
| Proportionality | \(A \propto B \iff A = kB\) (k is constant) | If distance is prop. to time, \(D = kt\) |
| Sum of Cubes | \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\) | \(x^3 + 8 = (x+2)(x^2 - 2x + 4)\) |
| Difference of Squares | \(a^2 - b^2 = (a-b)(a+b)\) | \(y^2 - 9 = (y-3)(y+3)\) |
Additional Information: Conditions and Special Cases
In our solution, we divided by \((a+b)\). This step is valid only if \((a+b) \neq 0\), which means \(a \neq -b\). If \(a = -b\), then \(a+b = 0\). In this case:
- \(a^3 + b^3 = a^3 + (-a)^3 = a^3 - a^3 = 0\)
- \(a^2 - b^2 = a^2 - (-a)^2 = a^2 - a^2 = 0\)
The initial condition \((a^3 + b^3) \propto (a^2 – b^2)\) becomes \(0 \propto 0\), which is true for any constant \(k\). The expression \((a^2 - ab + b^2)\) becomes \(a^2 - a(-a) + (-a)^2 = a^2 + a^2 + a^2 = 3a^2\).
And \((a-b)\) becomes \(a - (-a) = 2a\).
So, for \(a = -b\), we have \(3a^2\) proportional to \(2a\). This means \(3a^2 = k(2a)\). If \(a \neq 0\), then \(3a = 2k\), so \(k = \frac{3a}{2}\). The constant of proportionality \(k\) depends on \(a\) (or \(b\)), which contradicts the idea of \(k\) being a fixed constant for the general proportionality relationship.
Therefore, the proportionality \((a^2 - ab + b^2) = k(a-b)\) holds generally when \((a+b) \neq 0\). The question implicitly assumes the proportionality holds for generic values of \(a\) and \(b\) where the expressions are non-zero.
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