if a² -4a-1=0 find 3a³+3/a³
Answers
Answer:
f a + b + c = 5, a2 + b2 + c2 = 27, and a3 + b3 + c3 = 125
Step-by-step explanation:
Given :-
a²-4a-1 = 0
To find :-
Find the value of 3a³+(3/a³) ?
Solution :-
Given that
a²-4a-1 = 0
=> a²-(2×2×a) -1 = 0
=> a²-2(a)(2) -1 = 0
=> a²-2(a)(2) = 1
On adding 2² both sides then
=> a²-2(a)(2) +2²= 1+2²
=> (a-2)² = 1+4
=> (a-2)² = 5
=> a-2 = ±√5
=> a = 2±√5
Therefore, a = 2+√5 or 2-√5
If a = 2+√5 then 1/a = 1/(2+√5)
On Rationalising the denominator then
=> [1/(2+√5)]×[(2-√5)/(2-√5)]
=> (2-√5)/[(2+√5)(2-√5)]
=> (2-√5)/[2²-(√5)²]
Since (a+b)(a-b) = a²-b²
=>(2-√5)/(4-5)
=> (2-√5)/(-1)
=> -2+√5
=> 1/a = √5-2
Now,
We know that
(a+b)³ = a³+b³+3ab(a+b)
=> a³+b³ = (a+b)³-3ab(a+b)
Now,
a³+(1/a³) = [a+(1/a)]³-3(a)(1/a)[a+(1/a)]
=> a³+(1/a³) = (2+√5+√5-2)³-3(2+√5+√5-2)
=> a³+(1/a³) = (√5+√5)³-3(√5+√5)
=> a³+(1/a³) = (2√5)³-3(2√5)
=> a³+(1/a³) = 40√5-6√5
=> a³+(1/a³) = 34√5
Now
3a³+(3/a³) = 3[a³+(1/a³)]
=> 3(34√5)
=> 102√5
If a = 2-√5 then 1/a = 1/(2-√5)
On Rationalising the denominator then
=> [1/(2-√5)]×[(2+√5)/(2+√5)]
=> (2+√5)/[(2+√5)(2-√5)]
=> (2+√5)/[2²-(√5)²]
Since (a+b)(a-b) = a²-b²
=>(2+√5)/(4-5)
=> (2+√5)/(-1)
=> -2-√5
=> 1/a = -2-√5
Now,
We know that
(a+b)³ = a³+b³+3ab(a+b)
=> a³+b³ = (a+b)³-3ab(a+b)
Now,
a³+(1/a³) = [a+(1/a)]³-3(a)(1/a)[a+(1/a)]
=> a³+(1/a³) = (2-√5-2-√5)³-3(2-√5-2-√5)
=> a³+(1/a³) = (-√5-√5)³-3(-√5-√5)
=> a³+(1/a³) = (-2√5)³-3(-2√5)
=> a³+(1/a³) = -40√5+6√5
=> a³+(1/a³) = -34√5
Now
3a³+(3/a³) = 3[a³+(1/a³)]
=> 3(-34√5)
=> -102√5
Answer:-
The value of 3a³+(3/a³) is 102√5 or -102√5
Used formulae:-
→ The Rationalising factor of a+√b is a-√b
→(a+b)(a-b) = a²-b²
→ (a+b)³ = a³+b³+3ab(a+b)