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Question
If `9x^2 + 1/(4x^2) = 13,` find the value of `27x^3 + 1/(8x^3)`
Sum
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Solution
Given: `9x^2+1/(4x^2)=13,`
Let’s express it in terms of a binomial:
Here, a = 3x, b = `1/(2x),`
Then,
a2 = 9x2, b2 = `1/(4x^2)`
So, `(3x)^2 + (1/(2x))^2 = 13`
∴ a2 + b2 = 13
Finding a + b:
(a + b)2 = a2 + b2 + 2ab
(a + b)2 = 13 + 2`(3x)(1/(2x))`
(a + b)2 = 13 + 3
(a + b)2 = 16
a + b = `+-sqrt16`
∴ a + b = ±4
Using the identity,
a3 + b3 = (a + b)3 − 3ab(a + b)
a3 + b3 = (4)3 − 3`(3x)(1/(2x))`(4)
a3 + b3 = 64 − 3`(3/2)(4)`
a3 + b3 = 64 − `(9/2xx4)`
a3 + b3 = 64 − `(36/2)`
a3 + b3 = 64 − 18
∴ a3 + b3 = 46
Hence, `27x^3 + 1/(8x^3) = a^3 + b^3 = 46`
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Chapter 3: Expansions - EXERCISE B [Page 36]
