Advertisements
Advertisements
प्रश्न
Expand (ab – c)3
योग
Advertisements
उत्तर
(ab – c)3
Comparing (ab – c)3 with (a – b)3 we have a = ab and b = c
(a – b)3 = a3 – 3a2b + 3ab2 – b3
(ab – c)3 = (ab)3 – 3(ab)2c + 3ab(c)2 – c3
= a3b3 – 3(a2b2)c + 3abc2 – c3
= a3b3 – 3a2b2c + 3abc2 – c3
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Algebra - Exercise 3.3 [पृष्ठ ९१]
APPEARS IN
संबंधित प्रश्न
Expand.
(4 − p)3
Expand.
`(2p - 1/(2p))^3`
Expand.
`(1 - 1/a)^3`
Expand.
`(x/3 - 3/x)^3`
Expand (4p - 5q)3
Find the cube of 99 using the expansion formula.
Simplify.
(2x + 3y)3 - (2x - 3y)3
If `(y - 1/y)^3` = 27, then find the value of `y^3 - 1/y^3`
Expand (97xy)3
(a – b) = 3 and ab = 5 then a3 – b3 = __________
