Math, asked by ksamundee, 26 days ago

(i) (2x + 1)^3 (ii) (2a - 3b)^3 (iii) [3x/2 + 1]^3 (iv) [x - 2y/3]^3

Answers

Answered by looksneeraj33
1

Answer:

(ii) x – x3. (iii) y + y2 + 4. (iv) 1 + x. (v) 3t. ( vi) r2. (vii) 7x3. Ans. (i) x2 +x. ∵ The degree of x2 + x is 2. ∴ It is a quadratic ...

Answered by yashisharma14402
0

Step-by-step explanation:

(i) Using identity: (a + b)³ = a³+ b³+ 3ab(a + b)

(2x + 1)³ = (2x)³ + 1³ + (3×2x×1)(2x + 1)

= 8x³+ 1 + 6x(2x + 1)

= 8x³ + 1 + 12x² + 6x

(ii) Using identity, (a – b)³ = a³–b³ – 3ab(a – b)

(2a – 3b)³ = (2a)³– (3b)³ – (3×2a×3b)(2a – 3b)

=8a³–27b³–18ab(2a –3b)

= 8a³–27b³–36a²b + 54ab²

(iii) Using identity, (a + b)³ = a³+ b³+ 3ab(a + b)

[3x/2 +1]³ =(3x/2)³+1³+ (3×(3x/2)×1)(3x/2+ 1)

=27x³/8+1+9/2x×(3x/2+1)

= 27x³/8 + 1 + 27/4 x² + 9/2x

= (27/8)x³ + (27/4) x² + 9/2 x + 1

(iv) Using identity, (a – b)³=a³-b³-3ab(a-b) (X+ 2/3y)

= (x)³–(2/3 y)³– (3×x×2/3 y)(x – 2/3 y)

= x³– 8y³/27–2xy(x – 2/3 y)

= x³– (8/27)y³–2x²y+ 4/3xy²

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