1. Cubic Identities intro

2. Cubic Identities
Let us learn three important Cubic Identities which are very
useful in solving many problems.
We obtain these Identities by multiplying a same binomials
three times.

3. Identity 1 Let us consider (a + b)3
(a + b)3 = (a + b) (a + b) (a + b)
= [(a + b) (a + b)] (a + b)
= (a + b)2 (a + b)
= [a2+2ab+b2] (a + b)
= a (a2+2ab+b2) + b (a2+2a+b2)
= a3+2a2b +ab2 + ba2+2ab2+b3
= a3+2a2b +a2b+ ab2+2ab2+b3 = a3+3a2b + 3ab2+b3
Thus, (a + b)3 = a3+3a2b+3ab2+b3 x
We know that, (a + b)2 = a2+2ab+b2 x x x x x x x
By combining like terms

4. Identity 2 Let us consider (a - b)3
(a - b)3 = (a - b) (a - b) (a - b)
= [(a - b) (a - b)] (a - b)
= (a - b)2 (a - b)
= [a2- 2ab+b2] (a - b)
= a (a2- 2ab+b2) - b (a2- 2a+b2)
= a3 - 2a2b +ab2 - ba2+ 2ab2 - b3
= a3 -2a2b - a2b +ab2+2ab2 – b3 = a3-3a2b +3ab2 -b3
Thus, (a - b)3 = a3-3a2b+3ab2-b3 x x x x
We know that, (a - b)2 = a2-2ab+b2 x x x x
By combining like terms

5. Identity 3 Let us consider (x + a) (x + b) (x + c)
(x + a) (x + b) (x + c) = (x + a) (x + b) (x + c)
= [x2 + (a + b) x + ab] (x + c)
= x[x2 + (a + b) x + ab] + c(x2 + (a + b) x + ab)
= x3 + (a + b) x2 + abx + cx2 + (a + b) cx + abc
= x3 + ax2 + bx2 + abx + cx2 + acx + bcx + abc
= x3 + (a + b + c)x2 + (ab + bc + ca)x + abc
Thus, (x + a) (x + b) (x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc x
We know that, (x + a) (x + b)
= x2 + (a + b) x + ab x x x x x x x x x x x

6. Summary
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(a - b)3 = a3 - 3a2b + 3ab2 - b3
(x + a) (x + b) (x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc
Deduced Identities
a3 + b3 = (a + b) (a2 – ab + b2)
a3 - b3 = (a - b) (a2 + ab + b2)

7. Example 1:- Find the value of (x + 3)3 using suitable identity
Solution:
The above algebraic expression is same as following identity
(a + b)3 = a3 + 3a2b + 3ab2 + b3 Where a = x and b = 3
a3 + 3a2b + 3ab2 + b3
=x3 + 3 x x2 x x x x (After substituting value of a and b)
=x3 + 9x2 + 3 x 9 x x + 27
= x3 + 9x2 + 27x (Ans)

8. Example 2:- Find the value of (y - 5)3 using suitable identity
Solution:
The above algebraic expression is same as following identity
(a - b)3 = a3 - 3a2b + 3ab2 – b3 Where a = y and b = 5
a3 - 3a2b + 3ab2 – b3
=y3 - 3 x y2 x x y x (After substituting value of a and b)
=y y2 + 3 x 25 x y - 125
=y y2 + 75y (Ans)

9. The above algebraic expression is same as following identity
Example 3:- Find the value of (p + 3) (p + 5) (p + 4) using suitable identity
Solution:
The above algebraic expression is same as following identity
(x + a) (x + b) (x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc
x3 + (a + b + c)x2 + (ab + bc + ca)x + abc
=p3 + ( )x2 + (3 x x x 3)x + 3 x 5 x 4
= p3 + 12x2 + ( )x + 60
= p3 + 12x2 + 47x + 60 (Ans)
Where x = p, a = 3, b = 5 and c = 4
(After substituting values of x, a , b and c)

10. The above algebraic expression is same as following identity
Example 4:- Find the value of (m + 1) (m - 2) (m - 4) using suitable identity
Solution:
The above algebraic expression is same as following identity
(x + a) (x + b) (x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc
x3 + (a + b + c)x2 + (ab + bc + ca)x + abc
=m3 + (1 + (-2) + (-4))x2 + (1 x -2 + (-2) x (-4) + (-4) x 1)x + 1 x (-2) x (-4)
= m3 + (1 – 2 – 4)m2 + ( )m + 8
= m3 + (-5)m2 + 2m + 8 (Ans)
Where x = m, a = 1, b = -2 and c = -4
(After substituting values of x, a , b and c)
By rule, + (-) = -
By rule, (+) x (-) = - , (-) x (-) = +
Add rule 1, numbers having same sign , add and put the bigger number sign
Add rule 2, numbers having different sign , subtract and put the bigger number sign

11. Try these Using identity solve (p + 4)3 (x - 7)3
(m + 4) (m + 5) (m + 6)
(g – 3) (g – 5) (g + 7)