Question 32 of 43intermediate🔧 ApplyNumerical8 marks

Find the values using suitable identities:

(i)
17×2117 \times 21
Answer

Identity: a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b) Let a=19a = 19 and b=2b = 2.

17×21=(19−2)(19+2)=192−22=361−4=35717 \times 21 = (19 - 2)(19 + 2) = 19^2 - 2^2 = 361 - 4 = 357
(ii)
104×96104 \times 96
Answer

Identity: a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b) Let a=100a = 100 and b=4b = 4.

104×96=(100+4)(100−4)=1002−42=10000−16=9984104 \times 96 = (100 + 4)(100 - 4) = 100^2 - 4^2 = 10000 - 16 = 9984
(iii)
24×1624 \times 16
Answer

Identity: a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b) Let a=20a = 20 and b=4b = 4.

24×16=(20+4)(20−4)=202−42=400−16=38424 \times 16 = (20 + 4)(20 - 4) = 20^2 - 4^2 = 400 - 16 = 384
(iv)

1473147^{3}

Answer

Identity: (a+b)3=a3+3a2b+3ab2+b3(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 Let a=150a = 150 and b=−3b = -3.

(150−3)3=1503+3(150)2(−3)+3(150)(−3)2+(−3)3(150 - 3)^3 = 150^3 + 3(150)^2(-3) + 3(150)(-3)^2 + (-3)^3 =3375000−202500+4050−27=3176523= 3375000 - 202500 + 4050 - 27 = 3176523
(v)

1993199^{3}

Answer

Identity: (a+b)3=a3+3a2b+3ab2+b3(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 Let a=200a = 200 and b=−1b = -1.

(200−1)3=2003+3(200)2(−1)+3(200)(−1)2+(−1)3(200 - 1)^3 = 200^3 + 3(200)^2(-1) + 3(200)(-1)^2 + (-1)^3 =8000000−120000+600−1=7880599= 8000000 - 120000 + 600 - 1 = 7880599
(vi)

1273127^{3}

Answer

Identity: (a+b)3=a3+3a2b+3ab2+b3(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 Let a=120a = 120 and b=7b = 7.

(120+7)3=1203+3(120)2(7)+3(120)(7)2+73(120 + 7)^3 = 120^3 + 3(120)^2(7) + 3(120)(7)^2 + 7^3 =1728000+302400+17640+343=2048383= 1728000 + 302400 + 17640 + 343 = 2048383
(vii)

(-107)^3

Answer

Identity: (a+b)3=a3+3a2b+3ab2+b3(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

(−107)3=−(107)3(-107)^3 = - (107)^3

Let a=100a = 100 and b=7b = 7.

(100+7)3=1003+3(100)2(7)+3(100)(7)2+73(100 + 7)^3 = 100^3 + 3(100)^2(7) + 3(100)(7)^2 + 7^3 =1000000+210000+14700+343=1225043= 1000000 + 210000 + 14700 + 343 = 1225043

So, (−107)3=−1225043(-107)^3 = -1225043

(viii)

(-299)^3

Answer

Identity: (a+b)3=a3+3a2b+3ab2+b3(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

(−299)3=−(299)3(-299)^3 = - (299)^3

Let a=300a = 300 and b=−1b = -1.

(300−1)3=3003+3(300)2(−1)+3(300)(−1)2+(−1)3(300 - 1)^3 = 300^3 + 3(300)^2(-1) + 3(300)(-1)^2 + (-1)^3 =27000000−270000+900−1=26730899= 27000000 - 270000 + 900 - 1 = 26730899

So, (−299)3=−26730899(-299)^3 = -26730899

Explanation

For parts (i), (ii), and (iii), we use the identity a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b) to simplify the product of two numbers by expressing them as (a+b)(a+b) and (a−b)(a-b). For parts (iv) to (viii), we use the identity (a+b)3=a3+3a2b+3ab2+b3(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3. For negative numbers, we extract the negative sign first and then apply the identity to the positive counterpart, using a negative value for bb where suitable (e.g., 299=300−1299 = 300 - 1).

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