Advertisements
Advertisements
Question
Find the cube of : `2a + 1/(2a)` ( a ≠ 0 )
Sum
Solution
( a + b )3 = a3 + 3ab( a + b ) + b3
=`( 2a)^3 + (1/(2a))^3 = 3 xx cancel(2a) xx1/cancel(2a)( 2a + 1/(2a))`
= `8a^3 + 1/(8a^3) + 3 (2a + 1/(2a))`
= `8a^3 + 1/(8a^3) + 6a + 3/(2a)`
shaalaa.com
Is there an error in this question or solution?
APPEARS IN
RELATED QUESTIONS
Expand.
`((5x)/y + y/(5x))^3`
Simplify.
(3r − 2k)3 + (3r + 2k)3
Find the cube of : `( 3a - 1/a ) (a ≠ 0 )`
If a2 + `1/a^2 = 47` and a ≠ 0 find :
- `a + 1/a`
- `a^3 + 1/a^3`
If 4x2 + y2 = a and xy = b, find the value of 2x + y.
Expand : (3x + 5y + 2z) (3x - 5y + 2z)
If `"m"^2 + (1)/"m"^2 = 51`; find the value of `"m"^3 - (1)/"m"^3`
If `"r" - (1)/"r" = 4`; find : `"r"^3 - (1)/"r"^3`
Expand: (3x + 4y)3.
If `x^2 + 1/x^2` = 23, then find the value of `x + 1/x` and `x^3 + 1/x^3`
