Advertisements
Advertisements
Question
Find the value of `(sqrt(3) + 1)^4- (sqrt(3) - 1)^4`.
Solution
`(sqrt(3) + 1)^4 = ""^4"C"_0(sqrt(3))^4 (1)^0 + ""^4"C"_1(sqrt(3))^3 (1)^1 + ""^4"C"_2(sqrt(3))^2 (1)^2 + ""^4"C"_3(sqrt(3))^1 (1)^3 + ""^4"C"_4 (sqrt(3))^0 (1)^4`
Since, 4C0 = 4C4 = 1, 4C1 = 4C3 = 4,
4C2 = `(4 xx 3)/(2 xx 1)` = 6
∴ `(sqrt(3) + 1)^4 = 1(9)(1) + 4(3sqrt(3))(1) + 6(3) (1) + 4(sqrt(3))(1) + 1(1)(1)`
∴ `(sqrt(3) + 1)^4 = 9 + 12sqrt(3) + 18 + 4sqrt(3) + 1` ...(i)
Also, `(sqrt(3) - 1)^4 = ""^4"C"_0(sqrt(3))^4 (1)^0 - ""^4"C"_1(sqrt(3))^3 (1)^1 + ""^4"C"_2(sqrt(3))^2(1)^2 - ""^4"C"_3(sqrt(3))^1(1)^3 + ""^4"C"_4(sqrt(3))^0 (1)^4`
= `1(9)(1) - 4(3sqrt(3))(1) + 6(3)(1) - 4(sqrt(3))(1) + 1(1)(1)`
∴ `(sqrt(3) - 1)^4 = 9 - 12sqrt(3) + 18 - 4sqrt(3) + 1` ...(ii)
Subtracting (ii) from (i), we get
`(sqrt(3) + 1)^4- (sqrt(3) - 1)^4`
= `(9 + 12sqrt(3) + 18 + 4sqrt(3) + 1) - (9 - 12sqrt(3) + 18 - 4sqrt(3) + 1)`
= `24sqrt(3) + 8sqrt(3)`
= `32sqrt(3)`.
APPEARS IN
RELATED QUESTIONS
Expand: `(sqrt(3) + sqrt(2))^4`
Expand: `(sqrt(5) - sqrt(2))^5`
Expand: (2x2 + 3)4
Expand: `(2x - 1/x)^6`
Find the value of `(2 + sqrt(5))^5 + (2 - sqrt(5))^5`
Prove that `(sqrt(3) + sqrt(2))^6 + (sqrt(3) - sqrt(2))^6` = 970
Prove that `(sqrt(5) + 1)^5 - (sqrt(5) - 1)^5` = 352
Using binomial theorem, find the value of (102)4
Using binomial theorem, find the value of (1.1)5
Using binomial theorem, find the value of (9.9)3
Using binomial theorem, find the value of (0.9)4
Without expanding, find the value of (x + 1)4 − 4(x + 1)3 (x − 1) + 6 (x + 1)2 (x − 1)2 − 4(x + 1) (x − 1)3 + (x − 1)4
Without expanding, find the value of (2x − 1)4 + 4(2x − 1)3 (3 − 2x) + 6(2x − 1)2 (3 − 2x)2 + 4(2x − 1)1 (3 − 2x)3 + (3 − 2x)4
Find the value of (1.02)6, correct upto four places of decimal
Find the value of (1.01)5, correct up to three places of decimals.
Find the value of (0.9)6, correct upto four places of decimal
