Find the values using suitable identities: Class 9
Find the values using suitable identities: Class 9
Question 1.
Find the values using suitable identities: Class 9
i. 17 × 21
ii. 104 × 96
iii. 24 × 16
iv. 147³
v. 199³
vi. 127³
vii. (-107)³
viii. (-299)³
Solution:
i. 17 × 21 = (19 - 2)(19 + 2)
= 19² - 2² ...[∵ (a + b)(a - b) = a² - b²]
= 361 - 4 = 357
ii. 104 × 96
= (100 + 4)(100 - 4)
= 100² - 4² ...[∵ (a + b)(a - b) = a² - b²]
= 10000 - 16 = 9984
iii. 24 × 16 = (20 + 4)(20 - 4)
= 20² - 4² ...[∵ (a + b)(a - b) = a² - b²]
= 400 - 16 = 384
iv. 147³ = (150 - 3)³
= 150³ - 3(150)²(3) + 3(150)(3)² - 3³
... [∵ (a - b)³ = a³ - 3a²b + 3ab² - b³]
= 3375000 - 202500 + 4050 - 27
= 3176523
v. 199³ = (200 - 1)³
= 200³ - 3(200)²(1) + 3(200)(1)² - 1³
...[∵ (a - b)³ = a³ - 3a²b + 3ab² - b³]
= 8000000 - 120000 + 600 - 1
= 7880599
vi. 1273 = (130 - 3)³
= 130³ - 3(130)²(3) + 3(130)(3)² - 3³
...[∵ (a - b)³ = a³ - 3a²b + 3ab² - b³]
= 2197000 - 152100 + 3510 - 27
= 2048383
vii. (-107)³ = -(107)³
Now, (107)³ = (100 + 7)³
= 100³ + 3(100)²(7) + 3(100)(7)² + 7³
...[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= 1000000 + 210000 + 14700 + 343
=1225043
∴ (-107)³ = -1225043
viii. (-299)³ = -(299)³
Now,
(299)³ = (300 - 1)³
= 300³ - 3(300)²(1) + 3(300)(1)² - 1³ .....[∵ (a - b)³ = a³ - 3a²b + 3ab² - b³]
= 27000000 - 270000 + 900 - 1
= 26730899
∴ (-299)³ = -26730899