a3 b3 Formula or a3 b3 Formula is one of the most important and most asked formulae in Mathematics. Understanding the formula and applying it to solve the questions is easy. In this article, we explain the a3 b3 formula and a3 b3 c3 formula with proofs, along with solved questions, as well as unsolved for your practice.
a3 b3 Formula and a3 b3 Formula Proof
a3 b3 Formula is a polynomial equation with two variables. It is an equation in the form P=0, where P is the polynomial. A polynomial is a mathematical expression that solely uses the operations of subtraction, addition, multiplication, and positive-integer powers of variables. It is made up of indeterminates, also known as variables, such as x, y and z, and numeric coefficients.
The a3 b3 Formula are as follows.
a3 + b3 = (a + b)(a2 – ab + b2)
a3 – b3 = (a – b)(a2 + ab +b2)
Proof of a3 + b3 formula
As we know that:
(a + b)3 = a3 + b3 + 3ab(a + b)
Hence:
a3 + b3 = (a + b)3 – 3ab(a + b)
Taking (a + b) common in RHS, we get
a3 + b3 = (a + b)[(a + b)2 – 3ab]
a3 + b3 = (a + b)[(a2 + b2 + 2ab) – 3ab]
a3 + b3 = (a + b)(a2 + b2 + 2ab –3ab)
a3 + b3 = (a + b)(a2 – ab + b2)
Proof of a3 – b3 formula
As we know that:
(a – b)3 = a3 – b3 – 3ab(a – b)
Hence:
a3 – b3 = (a – b)3 + 3ab(a – b)
Taking (a – b) common in RHS, we get
a3 – b3 = (a – b)[(a – b)2 + 3ab]
a3 – b3 = (a – b)[(a2 + b2 – 2ab) + 3ab]
a3 – b3 = (a – b)(a2 + b2 – 2ab + 3ab)
a3 – b3 = (a – b)(a2 + b2 + ab)
See Also: a2 b2 formula
Solved Questions and Answers for a3 b3 Formula
In this section, we have provided a few solved questions that involve usage of a3 b3 Formula.
Q1) If a = 7 and b = 3,
Find the value of (a2 – ab + b2) using a3 b3 Formula
Ans)
Given: a = 7 and b = 3
As per the given data,
a + b = 10
And
a3 + b3 = 73 + 33
a3 + b3 = 343 + 27
a3 + b3 = 370
As we know that
a3 + b3 = (a + b)(a2 – ab + b2)
Substituting value of a3 + b3 and a + b:
370 = 10 X (a2 – ab + b2)
Hence
(a2 – ab + b2) = 370/10
(a2 – ab + b2) = 37
Practice questions
Ques 1) If a = 7 and ab = 28
Find the value of a3 – b3 using a3 b3 Formula
Ques 2) If a + b = 11 and ab = 18
Find the value of a3 + b3 using a3 b3 Formula
a3 b3 c3 Formula
a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
Proof of a3 b3 c3 Formula
The following is the proof of a3 b3 c3 formula
We take LHS
a3 + b3 + c3 – 3abc
Adding and subtracting 3ab(a + b)
a3 + b3 + 3ab(a + b) – 3ab(a + b) + c3 – 3abc
As we know that: (a + b)3 = a3 + b3 + 3ab(a + b)
[a3 + b3 + 3ab(a + b)] – 3ab(a + b) + c3 – 3abc
(a + b)3 – 3ab(a + b) + c3 – 3abc
Adding and subtracting 3(a + b)c(a + b + c)
(a + b)3 + c3 + 3(a + b)c(a + b + c) – 3(a + b)c(a + b + c) – 3ab(a + b) – 3abc
[(a + b)3 + c3 + 3(a + b)c(a + b + c)] – 3(a + b)c(a + b + c) – 3ab(a + b + c)
(a + b + c)3 – 3(a + b)c(a + b + c) – 3ab(a + b + c)
Taking (a + b + c) common
(a + b + c)[(a + b + c)2 – 3(a + b)c – 3ab]
(a + b + c)[(a + b + c)2 – 3(ab + bc + ca)]
As we know that: (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
(a + b + c)[a2 + b2 + c2 + 2(ab + bc + ca) – 3(ab + bc + ca)]
(a + b + c)[a2 + b2 + c2 – (ab + bc + ca)]
(a + b + c)(a2 + b2 + c2 – ab – bc – ca)
Hence Proved
Solved questions using a3 b3 c3 Formula
In this section, we have provided a few solved questions that involve usage of a3 b3 c3 Formula.
Ques) If a = 4, b = 7 and c = 1
Find the value of (a2 + b2 + c2 – ab – bc – ca) using a3 b3 c3 Formula
Ans) Given: a = 4, b = 7 and c = 1
Hence,
a + b + c = 12
a3 = 43 = 64
b3 = 73 = 343
c3= 13 = 1
abc = 4 X 7 X 1 = 28
a3 + b3 + c3 = 408
Substituting the values in the equation
a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
408 – 3 X 28 = 12 X (a2 + b2 + c2 – ab – bc – ca)
324 = 12 X (a2 + b2 + c2 – ab – bc – ca)
Hence,
(a2 + b2 + c2 – ab – bc – ca) = 324/12
(a2 + b2 + c2 – ab – bc – ca) = 27
Practice questions
Ques) If a = 7, b = 8, c = 9
Find the value of [a3 + b3 + c3– 3abc] using a3 b3 c3 formula
Thats it from our side
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