Simplification of (a + b + c)(a^2 + b^2 + c^2 – ab – bc – ca) with examples are available on this page. So, the students who are in search of (a + b + c)(a² + b² + c² – ab – bc – ca) simplification can utilize this article and learn the derivations. Practice the problems a number of times to make math as your favorite subject and become a master. Scroll down this page to see problems on Expansion of Powers of Binomials and Trinomials with a brief explanation.
Simplification of (a + b + c)(a^2 + b^2 + c^2 – ab – bc – ca) with Proof
(a + b + c)(a² + b² + c² – ab – bc – ca) = a(a² + b² + c² – ab – bc – ca) + b (a² + b² + c² – ab – bc – ca) + c(a² + b² + c² – ab – bc – ca)
(a + b + c)(a² + b² + c² – ab – bc – ca) = a³+ ab² + ac² – a²b – abc – a²c + a²b + b³ + c²b -ab² – b²c – abc + ca² + cb² + c³ – abc – bc² – c²a
(a + b + c)(a² + b² + c² – ab – bc – ca) = a³+ b³ + c³ – 3abc
Thus the simplification of (a + b + c)(a² + b² + c² – ab – bc – ca) is a³+ b³ + c³ – 3abc
Do Check:
Examples on Simplification of (a + b + c)(a^2 + b^2 + c^2 – ab – bc – ca)
Example 1.
Simplify the equation ( 3x + 3y + 3z) ( x² + 2y² + 3z² – 2xy -3yz – 4zx)
Solution:
Given that the equation is
( 3x + 3y + 3z) ( x² + 2y² + 3z² – 2xy -3yz – 4zx)
We know that
( a + b + c) ( a² + b² + c² – ab – bc – ca ) = a³ + b³ + c³ – 3abc
Therefore the given expression is (3x + 3y + 3z) {(3(3x)² + 3(3y)² + 3(3z)² – 2(3x)(3y) – 3(3y)(3z) – 4(3z)(3x)} = (3x)³ + (3y)³ + (3z)³ – 3(3x)(3y)(3z)
= 27x³ + 27x³ + 27x³ – 81
Example 2.
Simplify the equation ( x + 2y + z) ( x² + y² + z² – xy -yz – zx)
Solution:
Given that the equation is
( x + 2y + z) ( x² + y² + z² – xy -yz – zx)
We know that
( a + b + c) ( a² + b² + c² – ab – bc – ca ) = a³ + b³ + c³ – 3abc
Therefore the given expression is (x + 2y + z) {(x² + y² + z² – (x)(2y) – (2y)(z) – (z)(x)} = x³ + (2y)³ + z³ – 3x(2y)z
= x³ + 8y³ + z³ – 6xyz
Example 3.
If a + b + c = 16 and ab + bc + ca = 8 then the value of a³ + b³ + c³ – 3abc
Solution:
Given that
a + b + c = 16
ab + bc + ca = 8
We know that
( a + b + c) ( a² + b² + c² – ab – bc – ca ) = a³ + b³ + c³ – 3abc
= (16)(a² + b² + c² – 8)
Therefore a³ + b³ + c³ – 3abc = 16a² + 16b² + 16c² – 128
Example 4.
If a + b + c = 2 and ab + bc + ca = 8 then the value of a³ + b³ + c³ – 3abc
Solution:
Given that
a + b + c = 2
ab + bc + ca = 8
We know that
( a + b + c) ( a² + b² + c² – ab – bc – ca ) = a³ + b³ + c³ – 3abc
= (2)(a² + b² + c² – 8)
Therefore a³ + b³ + c³ – 3abc = 2a² + 2b² + 2c² – 16.
Example 5.
Simplify the equation (a + 2b + 4c) ( a² + b² + c² – 3ab + 6bc + 2ca)
Solution:
Given that the equation is
(a + 2b + 4c) (a² + b² + c² – 3ab + 6bc + 2ca)
We know that
( a + b + c) ( a² + b² + c² – ab – bc – ca ) = a³ + b³ + c³ – 3abc
Therefore the given expression is (a + 2b + 4c) {(a² + (2b)² + (4c)² – (a)(2b) – (2b)(4c) – (4c)(a)} = a³ + (2b)³ + (4c)³ – 3a(2b)(4c)
= a³ + 8b³ + 48c³ – 24abc
Thus the simplification of (a + 2b + 2c) ( a² + b² + c² – 2ab + 2bc + 2ca) is a³ + 8b³ + 48c³ – 24abc