Let us see the Special Expansions one by one.
Products
Sum of squares
Difference of squares
Sum of cubes
Difference of cubes
Products:
Let us first consider the product (a + b) (a + b) or (a + b)2.
(a + b)2 = (a + b) (a + b)
= a(a + b) + b (a + b)
= a2 + ab + ba + b2
= a2 + 2ab + b2
Thus,
(a + b)2 = a2 + 2ab + b2
One may verify that for any value of "a" and any value of "b", the values of the two sides are equal.
Find (2x + 3y)2
Solution:
(2x + 3y)2 = (2x)2 + 2(2x) (3y) + (3y)2 = 4x2 + 12xy + 9y2
Next we consider…
(a – b)2
= (a – b) (a – b)
= a (a – b) – b (a – b)
= a2 – ab – ba + b2
= a2 – 2ab + b2
or (a – b)2 = a2 – 2ab + b2
Find: (4p – 3q)2
Solution:
(4p – 3q)2
=(4p)2 – 2 (4p) (3q) + (3q)2
= 16p2 – 24pq + 9q2
Answer: (4p – 3q)2 = 16p2 – 24pq + 9q2
Now,
(a + b)3
= (a + b) (a + b)2
= (a + b) (a2 + 2ab+ b2)
= a(a2 + 2ab+ b2)+ b(a2 + 2ab+ b2)
= a3 + 2a2b + ab2 + a2b + 2ab2 + b3
= a3 + 3a2b + 3ab2 + b3
= a3 + b3 + 3ab(a + b)
Thus,
(a + b)3 = a3 + b3 + 3ab(a + b)
Also, by replacing "b" by "–b" in the above formula, we get
(a - b)3 = a3 - b3 - 3ab(a - b)
Write the following cubes in the expanded form:
(5p – 3q)3
Solution :
Comparing the given expression with (a – b)3, we find that
a = 5p, b = 3q.
We have,
(5p – 3q)3 = (5p)3 – (3q)3 – 3(5p)(3q)(5p – 3q)
= 125p3 – 27q3 – 225p2q + 135pq2
Example:
Evaluate (99)3
Solution :
(99)3
= (100 – 1)3
= (100)3 – (1)3 – 3(100)(1)(100 – 1)
= 100000 – 1 –29700
= 970299
Sum of Squares:
We know that,
(a + b)2 – 2ab
= (a2 + 2ab + b2) – 2ab
= a2 + 2ab + b2 – 2ab
= a2 + b2.
Thus, a2+b2 = (a+b)2 - 2ab
Also, (a – b)2 + 2ab
= (a2 – 2ab + b2) + 2ab
= a2 – 2ab + b2 + 2ab
= a2 + b2
Thus, a2+b2 = (a -b)2 + 2ab
Example:
If the values of a + b and ab are 12 and 32 respectively, find the values of
a2 + b2 and (a–b)2
Solution:
a2 + b2
= (a + b)2 –2ab
= (12)2 – 2(32)
= 144 – 64
= 80
a2 + b2 = 80
a2+b2 = (a -b)2 + 2ab
80 = (a -b)2 + 2(32)
80 = (a -b)2 + 64
16 = (a -b)2
Related Tags
Study of Special Expansions, What are Special Expansions, Sum of squares