a²+b²+c²−ab−bc−ca Formula Explained with Examples
The a²+b²+c²−ab−bc−ca formula rewrites the difference between the sum of three squares and their pairwise products. It can be expressed using squared differences or the sum and pairwise-product terms. These forms help calculate values, prove non-negativity and solve algebraic aptitude problems efficiently.
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What Is the a²+b²+c²−ab−bc−ca Formula?
The main identity is a²+b²+c²−ab−bc−ca = ½[(a−b)²+(b−c)²+(c−a)²]. Since every squared term is non-negative, the expression is always greater than or equal to zero. It is zero only when a = b = c. For example, when a = 4, b = 7 and c = 10, the value is 27.
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Three-Variable Square Identity
Learn the three-variable square identity and understand how to apply the expression A² + B² + C² − AB − BC − CA in algebraic simplification.
a²+b²+c²−ab−bc−ca Formula Formula & Tricks
Important Formulas
This form is obtained by adding the three squared differences and dividing by 2.
Use this form when the value of a+b+c and the value of ab+bc+ca are given.
This is the same identity without the fraction and is useful for direct expansion.
Quick Tricks
Find a−b, b−c and c−a, square them, add the results and divide by 2. This avoids separately calculating six terms.
If a+b+c and ab+bc+ca are known, apply (a+b+c)²−3(ab+bc+ca) directly.
a²+b²+c²−ab−bc−ca Formula Concepts
Derivation from squared differences
(a−b)²+(b−c)²+(c−a)² equals (a²−2ab+b²)+(b²−2bc+c²)+(c²−2ca+a²). Combining like terms gives 2a²+2b²+2c²−2ab−2bc−2ca. Dividing by 2 gives the required formula.
Non-negativity and equality condition
The pairwise-difference form contains only squared real numbers. The value equals zero exactly when all three squared differences are zero, which requires a=b, b=c and c=a. Hence, equality occurs only when a=b=c.
Using the sum and pairwise products
Since (a+b+c)²=a²+b²+c²+2(ab+bc+ca), subtracting 3(ab+bc+ca) leaves a²+b²+c²−ab−bc−ca. This method is useful when symmetric values are provided instead of individual variables.
Special case when a+b+c=0
Substituting c=−a−b into the expression gives 3(a²+ab+b²). Also, because (a+b+c)²=0, the sum form gives the expression as −3(ab+bc+ca). Thus, under this condition, ab+bc+ca is non-positive.
a²+b²+c²−ab−bc−ca Formula Data Table
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Equivalent forms of the expression
The same expression can be selected in different forms depending on the given values.
| Form | Identity | Best used when |
|---|---|---|
| Original form | a²+b²+c²−ab−bc−ca | The values of a, b and c are directly given |
| Difference form | ½[(a−b)²+(b−c)²+(c−a)²] | Pairwise differences are easy to calculate |
| Sum form | (a+b+c)²−3(ab+bc+ca) | The total sum and pairwise-product sum are given |
Note: All three forms have exactly the same value for real numbers a, b and c.
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Revision Points
Recall these identities and properties for quick algebraic calculations.
- a²+b²+c²−ab−bc−ca = ½[(a−b)²+(b−c)²+(c−a)²].
- a²+b²+c²−ab−bc−ca = (a+b+c)²−3(ab+bc+ca).
- The expression is non-negative for all real a, b and c.
- The expression equals zero if and only if a=b=c.
- If a+b+c=0, the expression becomes 3(a²+ab+b²) after substituting c=−a−b.
- Use pairwise differences when individual values are given and the sum form when symmetric information is given.
a²+b²+c²−ab−bc−ca Formula FAQs
What is the simplest form of a²+b²+c²−ab−bc−ca?
Its standard squared-difference form is ½[(a−b)²+(b−c)²+(c−a)²].
Is a²+b²+c²−ab−bc−ca always positive?
It is always non-negative for real a, b and c. It is zero when a=b=c; otherwise it is positive.
What is the value when a=3, b=5 and c=7?
The value is 9+25+49−15−35−21=12. Using differences, it is ½[(-2)²+(-2)²+4²]=12.
How do you calculate the expression if a+b+c and ab+bc+ca are known?
Use (a+b+c)²−3(ab+bc+ca). For a+b+c=10 and ab+bc+ca=21, the value is 100−63=37.
When does a²+b²+c²−ab−bc−ca equal zero?
It equals zero if and only if a=b=c, because all three pairwise squared differences must be zero.
What is the formula when a+b+c=0?
Putting c=−a−b gives a²+b²+c²−ab−bc−ca=3(a²+ab+b²). It is also equal to −3(ab+bc+ca) under this condition.