Question

(3x +1)(9x2 - 3x +1) - (3x - 1)(9x² + 3x +1)
solve with a³-b³ or a³+b³​

Answer 1

Answer:

(i) 3x + (1/3x) = 3

⇒ (3x + (1/3x))2 = (3)2

⇒ (3x)2 + (1/3x)2 + 2 × 3x × (1/3x )= 9

⇒ 9x2 + (1/9x2) + 2 = 9 ⇒ 9x2 + (1/9x2) =

 9 -2

⇒ 9x2 + (1/9x2 ) =  7

(ii) 3x + (1/3x) = 3

⇒ (3x + (1/3x))3 = (3)3

⇒ (3x)3 + (1/3x)3 + 3 × 3x × (1/3x) (3x +

.1/3x) = 27

⇒ 27x3 + (1/27x3) + 3(3x + (1/2x)) = 27

⇒ 27x3 + (1/27x3) + 3(3) = 27

⇒ 27x3 + (1/27x3) = 27 -9

⇒ 27x3 + (1/27x3) = 18.

Answer 2

$\Large\mathtt\green{ }\huge\underline\mathtt\red{Answer : }$

(i) 3x + (1/3x) = 3

⇒ (3x + (1/3x))2 = (3)2

⇒ (3x)2 + (1/3x)2 + 2 × 3x × (1/3x )= 9

⇒ 9x2 + (1/9x2) + 2 = 9 ⇒ 9x2 + (1/9x2) =

 9 -2

⇒ 9x2 + (1/9x2 ) =  7

(ii) 3x + (1/3x) = 3

⇒ (3x + (1/3x))3 = (3)3

⇒ (3x)3 + (1/3x)3 + 3 × 3x × (1/3x) (3x +

.1/3x) = 27

⇒ 27x3 + (1/27x3) + 3(3x + (1/2x)) = 27

⇒ 27x3 + (1/27x3) + 3(3) = 27

⇒ 27x3 + (1/27x3) = 27 -9

⇒ 27x3 + (1/27x3) = 18.

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