If x = -2 and y = 1, by using an identity find the value of the following:
(i) (4y²–9x²) (16y⁴+36x²y²+81x⁴)
(ii) (2/x - x/2)(4/x² + x²/4 + 1)
(iii)(5y + 15/y)(25y² - 75 + 225/y²)
Answers
Concept :
By using an identity :
a³ + b³ = (a + b)(a² + b² - ab)
a³ - b³ = (a - b)(a² + b² + ab)
Given:
(i) (4y² - 9x²) (16y⁴ + 36x²y² + 81x⁴)
= (4y² - 9x²) [(4y²)² + 9x² × 4y² + (9x²)²)
= (4y²)³ - (9x²)³
= 64y⁶ - 729x⁶
On putting x = -2 and y = 1 , we obtain :
= 64(1)⁶ - 729(-2)⁶
= 64 - 729 × 64
= 64 - 46656
= - 46592
Hence, the value of (4y² - 9x²) (16y⁴ + 36x²y² + 81x⁴) is - 46592
(ii) (2/x - x/2)(4/x² + x²/4 + 1)
= (2/x - x/2)[(2/x)² + (x/2)² + 2/x × x/2]
= (2/x)³ - (x/2)³
= 8/x³ - x³/8
On putting x = -2 , we obtain :
= 8/(-2)³ - (-2)³/8
= 8/-8 - (-8/8)
= - 1 + 1
= 0
Hence, the value of (2/x - x/2)(4/x² + x²/4 + 1) is 0.
(iii) (5y + 15/y)(25y² - 75 + 225/y²)
= (5y + 15/y)[(5y)² - 5y × 15/y + (15/y)²]
= (5y)³ + (15/y)³
= 125y³ + 3375/y³
On putting y = 1 , we obtain :
= 125 (1)³ + 3375/(1)³
= 125 + 3375
= 3500
Hence, the value of (5y + 15/y)(25y² - 75 + 225/y²) is 3500.
HOPE THIS ANSWER WILL HELP YOU……
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Step-by-step explanation:
(2/x - x/2)(4/x² + x²/4 + 1)
= (2/x - x/2)[(2/x)² + (x/2)² + 2/x × x/2]
= (2/x)³ - (x/2)³
= 8/x³ - x³/8
On putting x = -2 , we obtain :
= 8/(-2)³ - (-2)³/8
= 8/-8 - (-8/8)
= - 1 + 1
= 0