The value of (x - y)(x + y) + (y - z)(y + z) + (z - x)(z + x) is:
0
Evaluating the Algebraic Expression (x - y)(x + y) + ...
The question asks for the value of the expression: \((x - y)(x + y) + (y - z)(y + z) + (z - x)(z + x)\) To solve this, we can use a fundamental algebraic identity known as the 'difference of squares'.
Applying the Difference of Squares Identity
The difference of squares identity states that for any two terms 'a' and 'b': \(a^2 - b^2 = (a - b)(a + b)\) We can apply this identity to each part of the given expression:
- For the first term, let \(a=x\) and \(b=y\). Applying the identity, we get: \((x - y)(x + y) = x^2 - y^2\)
- For the second term, let \(a=y\) and \(b=z\). Applying the identity, we get: \((y - z)(y + z) = y^2 - z^2\)
- For the third term, let \(a=z\) and \(b=x\). Applying the identity, we get: \((z - x)(z + x) = z^2 - x^2\)
Simplifying the Expression
Now, we substitute these results back into the original expression:
\( (x^2 - y^2) + (y^2 - z^2) + (z^2 - x^2) \)Let's rearrange the terms to group the like terms together:
\( x^2 - x^2 + y^2 - y^2 + z^2 - z^2 \)Now, we can perform the subtraction for each pair of like terms:
- \(x^2 - x^2 = 0\)
- \(y^2 - y^2 = 0\)
- \(z^2 - z^2 = 0\)
Adding these results together:
\( 0 + 0 + 0 = 0 \)Therefore, the value of the expression \((x - y)(x + y) + (y - z)(y + z) + (z - x)(z + x)\) is 0.
Conclusion
By applying the difference of squares identity and simplifying the terms, we find that the entire expression evaluates to 0.
Similar Questions
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Important Questions from Algebric Equations
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