How to Apply the a³+b³+c³−3abc Identity?
a³+b³+c³−3abc identity expresses the difference between the sum of three cubes and three times their product as a product involving a+b+c. This page covers its factorised forms, the special case a+b+c=0, useful shortcuts, and numerical examples for solving algebra problems in quantitative aptitude.
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What is the a³+b³+c³−3abc Identity?
The second factor measures the differences among a, b and c. Since it equals half the sum of three squared differences, it is always non-negative for real values. For example, when a=1, b=2 and c=3, the left side is 1+8+27−18=18, while the factorised form gives 6(14−2−6−3)=6×3=18.
a³+b³+c³−3abc Identity Video Lessons
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Three-Variable Cubic Identity A³+B³+C³
Learn the algebraic identity for a³ + b³ + c³ and understand how it relates to the product term and the sum of the variables.
a³+b³+c³−3abc Identity Formula & Tricks
Important Formulas
Use this form when the sum a+b+c and the pairwise products are easy to calculate.
This follows from a²+b²+c²−ab−bc−ca = ½[(a−b)²+(b−c)²+(c−a)²].
The factor a+b+c becomes zero in the main factorisation.
Quick Tricks
If a+b+c=0, the entire expression a³+b³+c³−3abc is immediately zero. There is no need to calculate all three cubes.
Calculate the sum first and then use half the sum of squared differences. This can be shorter than evaluating three cubes.
If a=b=c, then both a+b+c and the squared-difference factor are valid, but all differences are zero. Therefore, a³+b³+c³−3abc=0.
a³+b³+c³−3abc Identity Concepts
Factorisation by the sum of three variables
The identity is obtained by multiplying (a+b+c)(a²+b²+c²−ab−bc−ca). The mixed terms cancel, leaving a³+b³+c³−3abc.
Equivalent squared-difference form
Using (a−b)²+(b−c)²+(c−a)² = 2(a²+b²+c²−ab−bc−ca), the identity becomes ½(a+b+c)[(a−b)²+(b−c)²+(c−a)²].
Case when a+b+c=0
This result follows directly from the factorised identity. It applies to positive, negative or zero real values as long as their sum is zero.
Condition for equality of the cubes and product
From the identity, equality requires (a+b+c)(a²+b²+c²−ab−bc−ca)=0. The second factor is zero only when a=b=c, because it is half the sum of squared differences.
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a³+b³+c³−3abc Identity Revision Points
Use these formulas and conditions for quick revision.
- a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca).
- The second factor equals ½[(a−b)²+(b−c)²+(c−a)²].
- If a+b+c=0, then a³+b³+c³=3abc.
- If a=b=c, then a³+b³+c³−3abc=0.
- For real values, a²+b²+c²−ab−bc−ca is always non-negative.
- Do not confuse a³+b³+c³−3abc with a³+b³+c³; the term 3abc must be subtracted before applying the identity.
a³+b³+c³−3abc Identity FAQs
What is the formula for a³+b³+c³−3abc?
a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca). It is also equal to ½(a+b+c)[(a−b)²+(b−c)²+(c−a)²].
What happens when a+b+c=0 in this identity?
The expression becomes zero: a³+b³+c³−3abc=0. Therefore, a³+b³+c³=3abc.
Evaluate 2³+3³+(−5)³−3(2)(3)(−5).
Since 2+3−5=0, the expression is 0. Directly, 8+27−125+90=0.
How is the squared-difference form derived?
Expand the squared differences: (a−b)²+(b−c)²+(c−a)² = 2(a²+b²+c²−ab−bc−ca). Substituting this into the main identity gives the equivalent form.
When is a³+b³+c³ equal to 3abc for real a, b and c?
It is equal when a+b+c=0 or when a=b=c. These are the two ways the product of the factors in the identity can be zero.
Find a³+b³+c³−3abc for a=4, b=5 and c=6.
Using the squared-difference form, the value is ½(15)[(4−5)²+(5−6)²+(6−4)²] = ½(15)(1+1+4)=45.