If ab + bc + ca = 0, then what is the value of a 2/(a 2– bc) + b 2/(b 2– ca) + c 2/(c 2– ab)?
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Evaluating an Algebraic Expression Given a Condition
We are asked to find the value of the expression \( \frac{a^2}{a^2 – bc} + \frac{b^2}{b^2 – ca} + \frac{c^2}{c^2 – ab} \) given the condition \( ab + bc + ca = 0 \). To evaluate this expression, we need to use the given condition to simplify the terms.
Simplifying the Denominators Using the Condition \(ab + bc + ca = 0\)
Let's work with the given condition \(ab + bc + ca = 0\). We can rearrange this equation to express \(bc\), \(ca\), and \(ab\) in terms of the other variables:
- From \(ab + bc + ca = 0\), we get \(bc = -ab - ca\). We can factor out 'a' to write \(bc = -(ab + ca) = -a(b+c)\).
- Similarly, \(ca = -ab - bc = -(ab + bc)\). We can factor out 'b' to write \(ca = -b(a+c)\).
- And \(ab = -bc - ca = -(bc + ca)\). We can factor out 'c' to write \(ab = -c(b+a)\).
Now, let's substitute these relationships into the denominators of the expression we want to evaluate:
- The first denominator is \(a^2 - bc\). Substituting \(bc = -ab - ca\): \[ a^2 - bc = a^2 - (-ab - ca) = a^2 + ab + ca \] We can factor 'a' from this expression: \[ a^2 + ab + ca = a(a + b + c) \]
- The second denominator is \(b^2 - ca\). Substituting \(ca = -ab - bc\): \[ b^2 - ca = b^2 - (-ab - bc) = b^2 + ab + bc \] We can factor 'b' from this expression: \[ b^2 + ab + bc = b(b + a + c) \]
- The third denominator is \(c^2 - ab\). Substituting \(ab = -bc - ca\): \[ c^2 - ab = c^2 - (-bc - ca) = c^2 + bc + ca \] We can factor 'c' from this expression: \[ c^2 + bc + ca = c(c + b + a) \]
Substituting Simplified Denominators Back into the Expression
Now we replace the original denominators with these simplified forms in the expression:
\[ \frac{a^2}{a(a + b + c)} + \frac{b^2}{b(a + b + c)} + \frac{c^2}{c(a + b + c)} \]Assuming that \(a, b, c\) are non-zero and \(a+b+c \neq 0\), we can cancel out one 'a' from the first term, one 'b' from the second term, and one 'c' from the third term:
\[ \frac{\cancel{a} \cdot a}{\cancel{a}(a + b + c)} + \frac{\cancel{b} \cdot b}{\cancel{b}(a + b + c)} + \frac{\cancel{c} \cdot c}{\cancel{c}(a + b + c)} \]This simplifies the expression to:
\[ \frac{a}{a + b + c} + \frac{b}{a + b + c} + \frac{c}{a + b + c} \]Combining the Terms with a Common Denominator
All the terms in the simplified expression now have the same denominator, which is \((a + b + c)\). We can combine the numerators over this common denominator:
\[ \frac{a + b + c}{a + b + c} \]Determining the Final Value
The expression simplifies to a fraction where the numerator is identical to the denominator. Assuming that the denominator \((a + b + c)\) is not equal to zero, the value of this fraction is 1.
Thus, under the given condition \(ab + bc + ca = 0\), the value of the expression \( \frac{a^2}{a^2 – bc} + \frac{b^2}{b^2 – ca} + \frac{c^2}{c^2 – ab} \) is 1.
Revision Table: Steps to Evaluate the Algebraic Expression
| Step | Action | Intermediate Result / Explanation |
|---|---|---|
| 1 | Start with the given expression and condition. | Expression: \( \frac{a^2}{a^2 – bc} + \frac{b^2}{b^2 – ca} + \frac{c^2}{c^2 – ab} \) Condition: \( ab + bc + ca = 0 \) |
| 2 | Use the condition to rewrite \(bc\), \(ca\), and \(ab\). | \( bc = -ab - ca \), \( ca = -ab - bc \), \( ab = -bc - ca \) |
| 3 | Substitute these into the denominators. | \( a^2 - bc = a(a+b+c) \) \( b^2 - ca = b(a+b+c) \) \( c^2 - ab = c(a+b+c) \) |
| 4 | Substitute simplified denominators into the expression. | \( \frac{a^2}{a(a+b+c)} + \frac{b^2}{b(a+b+c)} + \frac{c^2}{c(a+b+c)} \) |
| 5 | Simplify each term by cancelling common factors (assuming \(a,b,c, a+b+c \neq 0\)). | \( \frac{a}{a+b+c} + \frac{b}{a+b+c} + \frac{c}{a+b+c} \) |
| 6 | Combine terms with the common denominator. | \( \frac{a+b+c}{a+b+c} \) |
| 7 | Simplify the final fraction. | \( 1 \) |
Additional Information: Algebraic Simplification and Conditions
This problem is a classic example of how a given condition can be used to simplify a complex algebraic expression dramatically. The key insight is to see how the condition \(ab+bc+ca=0\) allows us to factor the denominators into a common form involving \((a+b+c)\).
The process of cancelling terms assumes that the terms being cancelled are not zero. Specifically, for the final step where \(\frac{a+b+c}{a+b+c}\) becomes 1, we assume that \(a+b+c \neq 0\). If \(a+b+c = 0\), the denominators \(a(a+b+c)\), \(b(a+b+c)\), \(c(a+b+c)\) would be zero, making the original terms undefined, unless the numerators were also zero in a way that allowed for limits or other interpretations (which is not the case here as seen from the numerators \(a^2, b^2, c^2\)). Given the multiple-choice options are specific numbers, the problem is set up such that the expression evaluates to a constant value under the general condition, implying the case where the simplification holds is the intended one.
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