Question

Evaluate the following using suitable identities. (i) (2x+1)³ (ii) (2a-3b)³ (iii) (x+2y+42)² (iv) (102) (v) (998)¹​

Answer 1

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We know, 

(a+b+c)2=a2+b2+c2+2ab+2bc+2ac

i) (x+2y+4z)2

=x2+(2y)2+(4z)2+2(x)(2y)+2(2y)(4z)+2(x)(4z)

=x2+4y2+16z2+4xy+16yz+8xz

ii) (2x−y+z)2

=(2x)2+(−y)2+(z)2+2(2x)(−y)+2(−y)(z)+2(2x)(z)

=4x2+y2+z2−4xy−2yz+4xz

iii) (−2x+3y+2z)2

=(−2x)2+(3y)2+(2z)2+2(−2x)(3y)

Answer 2

Step-by-step explanation:

(x + y)³ = x³ + y³ + 3xy(x + y)

(x - y)³ = x³ - y³ - 3xy(x - y)

i) (2x + 1)³

Identity: (x + y)³ = x³ + y³ + 3xy(x + y)

Here x = 2x, y = 1

(2x +1)³ = (2x)³ + (1)³ + 3(2x)(1)(2x + 1)

= 8x³ + 1 + 6x(2x + 1)

= 8x³ + 1 + 12x² + 6x

= 8x³ + 12x² + 6x +1

ii) (2a - 3b)³

Identity: (x - y)³ = x³ - y³ - 3xy(x - y)

Here x = 2a, y = 3b

(2a - 3b)³ = (2a)³ - (3b)³ - 3(2a)(3b)(2a - 3b)

= 8a³ - 27b3 - 18ab(2a - 3b)

= 8a³ - 27b³ - 36a²b + 54ab²

= 8a³ - 36a²b + 54ab² - 27b³

(x+y+z) 2 =x 2 +y 2 +z 2+2xy+2yz+2xz

Therefore, the value of (x+2y+4z) 2 is=

x 2 +(2y) 2 +(4z) 2

+2(x)(2y)+2(2y)(4z)+2(x)(4z)

=x 2 +4y 2 +16z 2 +4xy+16yz+8xz