A cube plus B cube plus C cube 3abc Ka Formula
What is the formula of sum of a3 b3 c3? and What is the identity of a3 b3 c3 3abc?. A cube plus B cube plus C cube 3abc Ka Formula equal to a3 + b3 + c3 = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) + 3abc. Which is an algebraic identity used to find the sum of cubes of the three numbers. The expression of this formula is a3 + b3 + c3 = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) + 3abc. Read about Factorization of the form (a3 + b3 + c3 – 3abc) and if a+b+c = 0 then a3+b3+c3 is equal to 3abc.
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Factorization of the form (a3 + b3 + c3 – 3abc) formula
Lets understand this a3 + b3 + c3 Formula in details or What is the identity of a3 b3 c3 3abc?. What is the formula of sum of A cube B Cube C Cube? This formula is the sum of the cubes of three numbers.
Algebraic Identity: a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 -ab – bc – ca). How is this algebraic identity obtained? Let’s see how.
Taking RHS of the above algebraic identity: (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
- Multiply each term of first polynomial with every term of second polynomial, as shown below:
- = a(a2 + b2 + c2 – ab – bc – ca ) + b(a2 + b2 + c2 – ab – bc – ca ) + c(a2 + b2 + c2 – ab – bc – ca )
- = { (a * a2) + (a * b2) + (a * c2) – (a * ab) – (a * bc) – (a * ca) } + {(b * a2) + (b * b2) + (b * c2) – (b * ab) – (b * bc) – (b * ca)} + {(c * a2) + (c * b2) + (c * c2) – (c * ab) – (c * bc) – (c * ca)}
- Now, solve multiplication in curly braces and we get:
- = a3 + ab2 + ac2 – a2b – abc – a2c + a2b + b3 + bc2 – ab2 – b2c – abc + a2c + b2c + c3 – abc – bc2 – ac2
- Rearrange the terms and we get:
- = a3 + b3 + c3 + a2b – a2b + ac2– ac2 + ab2 – ab2 + bc2 – bc2 + a2c – a2c + b2c – b2c – abc – abc – abc
- Above highlighted like terms will be subtracted and we get:
- = a3 + b3 + c3 – abc – abc – abc
- adding like terms i.e (-abc) and we get:
- = a3 + b3 + c3 – 3abc
- Hence, a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
List of Some important Algebraic formula
- (x + y + z)2 = x2 + y2 + z2 + 2xy +2yz + 2xz
- (x + y – z)2 = x2 + y2 + z2 + 2xy – 2yz – 2xz
- (x – y + z)2 = x2 + y2 + z2 – 2xy – 2yz + 2xz
- (x – y – z)2 = x2 + y2 + z2 – 2xy + 2yz – 2xz
- x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz -xz)
- x3 + y3 = (x + y) (x2 – xy + y2)
- x3 – y3 = (x – y) (x2 + xy + y2)
- (a – b)3 = a3 – b3 – 3ab (a – b)
- (a + b)3 = a3 + b3 + 3ab (a + b)
- (a + b)2 = a2 + 2ab + b2
- (a – b)2 = a2 – 2ab + b2
- (a + b) (a – b) = a2 – b2
- (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
- (a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3 + b4
Solved examples using by A cube plus B cube plus C cube 3abc formula
Now you can understand the formula of Algebraic Identity A cube plus B cube plus C cube 3abc Ka Formula a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 -ab – bc – ca) in detail by solving some good examples
Example 1: Solve 8a3 + 27b3 + 125c3 – 90abc = ?
Solution: First you should know A cube plus B cube plus C cube 3abc Ka Formula. a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 -ab – bc – ca). Given expression (8a3 + 27b3 + 125c3 – 90abc) can be written as:
(2a)3 + (3b)3 + (5c)3 – 3(2a)(3b)(5c) And this can be compared with a3 + b3 + c3 – 3abc
From above line we can write a = 2a, b = 3b and c = 5c
Now we can put these value of a, b and c in RHS of identity a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) and we get:
- = (2a + 3b + 5c) { (2a)2 +(3b)2 + (5c)2 – (2a)(3b) – (3b)(5c)– (5c)(2a) }
- Expand the exponential forms and we get:
- = (2a + 3b + 5c) {4a2 +9b2 + 25c2 – (2a)(3b) – (3b)(5c) – (5c)(2a)}
- Solve brackets and we get:
- = (2a + 3b + 5c) { 4a2 +9b2 + 25c2 – 6ab – 15bc – 10ca}
- Hence, 8a3 + 27b3 + 125c3 – 90abc = (2a + 3b + 5c) (4a2 +9b2 + 25c2 – 6ab – 15bc – 10ca)
FAQs on a b c whole cube formula
a3+b3+c3 = 3abc if a+b+c = 0. Formula – a3 + b3 + c3 = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) + 3abc
a3 + b3 + c3 = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) + 3abc. You can rewrite this formula as a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca).
a3 + b3 + c3 = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) + 3abc
a3 + b3 + c3 = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) + 3abc
a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
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