factorise 1+x³+1÷x³-3
Answers
Answer:
As we see that the x has a power of 3 so we begin by cubbing the whole equation.
So we start solving like:
x+1x=3
(x+1x)3=33
x3+3.x2.1x+3.x.1x2+1x3=27
(x3+1x3)+3.x2.1x+3.x.1x2=27
x3+1x3+(3x+3x)=27
x3+1x3+3(x+1x)=27
x3+1x3+3×3=27
x3+1x3+9=27
x3+1x3=18
Therefore the answer is 18
If x+1/x=2, then what is the value of x^3+1/x^3?
If x³+1/x³ =110,then does x+1/x=?
x+1x=3 then x5+1x5= ?
If x+1/x=3,then what will be the value of x?
If x+1/x=3, then what is the value of x^4+1/x^4=?
Due to Parenthetical Shifts in the proper Metric Form it will not resolve as notated moreso due to facts it associates to x+(1/x)=2.5 resolve for x=.5=2
Scientific Processes toward Deduction
From the getgo we observe two Resolves reliant on If you are Plainly Forgetting Brackets, we thus derive the following as Fig. 1 to 3:
(x+1)=3x resolves with easiness as: 2x=1=2×.5
x2+1=3x however, we observe this does not atall result same easiness of resolve as 1, and we get to Figure 3 as: (Example1×(x OR 1)) To share that Same Resolve as Figure 1 must also equal 3x(y OR 1) with something more alike 2xy+x=
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Answer will be 18.
If (x + 1/x) = 4 and (x^2 + 1/x^3) = 12, then what is the value of (x^3 + 1/x^2)?
If x³-1/x³=14,what is the value of x-1/x?
What is the solution of- if x+1/x=1, then (x+1) ^3+1/(x+1) ^3=?
If x3+1/x3=18 , then what is the value of x−1/x ?
If x^3+1/x^3=18 then what is x^4+1/x^4?
x+1/x =3
square both the sides-
x²+1/x² +2 ×x× 1/x =3²
x²+1/x² +2=9
x²+1/x² =7
Now to find value of x³+1/x³ first expand it
x³+1/x³ = (x+1/x) (x²+1/x² -1)
=(3)(7–1)
=3×6
x³+1/x³ =18
Given: x+1/x = 3
Cubing both sides
(x+1/x)^3 = 3^3
Now, using the formula ( a+b)^3
x^3 + 1/x^3 + 3× x × 1/x ( x+ 1/x) = 27
x^3 + 1/x^3 + 3 (x+1/x) = 27
A/q: x+ 1/x = 3
x^3 + 1/x^3 + 3 (3) = 27
x^3 + 1/x^3 = 27-9
x^3 + 1/x^3 = 18
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We need to find,
We are given,
We know that,
Lets take the cube of both sides of the given equation,
=> (x + 1/x)^3 = 3^3
=> x^3 + 1/x^3 + {3 . x . 1/x (x + 1/x)} = 27
=> x^3 + 1/x^3 + 3(x + 1/x) = 27
Since x + 1/x = 3
=> x^3 + 1/x^3 + 3×3 = 27
=> x^3 + 1/x^3 = 27-9
=> x^3 + 1/x^3 = 18
Here
(x+1/x)³=(3)³ (squaring both side)
By using identity (a+b)³=a³+b³+3ab (a+b)
(x)³+(1/x)³=27
x³+1/x³ +3×x×1/x (x+1/x)=27
x³+1/x³+3 (3)=27
x³+1/x³+9=27
x³+1/x³=27-9
x³+1/x³=18
X+1/x=3
Cubing on both sides
(X+1/x)^3=3^3
X^3+1/x^3+3×x×1/x(x+1/x)=27
X^3+1/x^3+3(x+1/x)=27
Since x+1/x=3
X^3+1/x^3+3×3=27
X^3+1/x^3+9=27
X^3+1/x^3=27–9
X^3+1/x^3=18
(x+1/x)^3=27
=>x^3+1/x^3+3(x)(1/x)[x+1/x]=27 {Using the formula of (x+y)^3}
=>using the value of x+1/x….
=>x^3+1/x^3+3*3=27
=>Or x^3+1/x^3=27–9=18.
x+(1/x)=3 , or, {x+(1/x)}^3=27
or, x^3+3x^2×(1/x)+3x×(1/x)^2 +(1/x)^3 =27
or, x^3+1/x^3+3x+3/x=27
or, x^3+1/x^3+3{(x+1)/x}=27
or, x^3 +1/x^3 +3×3=27
or, x^3 +1/x^3=27–9=18
Given , x+1/x =3
We know a^3 +b^3 = (a+b)^3 - 3ab(a+b) , so,
x^3 + 1/x^3
=(x)^3 + (1/x)^3
= (x+1/x)^3 - 3x*1/x(x+1/x)
=(3)^3 -3*1*3 { as x+1/x =3}
=27 - 9
= 18
x+1/x=3
x^5+1/x^5=(x^3+1/x^3)(x^2+1/x^2)-(x+1/x)
={(x+1/x)^3–3*x*1/x(x+1/x)}{(x+1/x)^-2*x*1/x)}-3
={(3)^3–3*3}{(9)^2–2)}-3
=(27–9)(9–2)-3
=18*7–3
=123(ans)
If x + 1/x = 3, then
x^2–3x+1 = 0
x1 = [3+(9-4)^0.5]’2 = [3+5^0.5]/2 = 2.618033989
(x1)^3 = 17.94427191
Hence x^3 + 1/(x^3) = 17.94427191 + 1/17.94427191 = 18.
Answer = 18.
If x+1/x=P then x³+1/x³=p³-3p
Here p=3, so x³+1/x³=3³-3×3=27–9=18.
Answer is 18.
Given,
X+1/X=3
cubing on both side,
(X+1/X)^3=3^3
X^3+1/X^3+3X*1/X(X+1/X)=27
X^3+1/X^3+3*3=27
X^3+1/X^3+9=27
X^3+1/X^3=27–9
X^3+1/X^3=18
18
x+1/x=3
=>(x+1/x)^3 = 3*3*3
=> x^3 + 1/x^3 + 3 * x * 1/x * (x + 1/x) = 27
=> x^3 + 1/x^3 = 27 - 3 * 3
=> x^3 + 1/x^3 = 18
x³ + 1/x³
= (x+1/x)³-3x*1/x(x+1/x)
= 3³-3*1*3[as (x+1/x)=3]
= 27 - 9 = 18
The value is 18 .
Ans is 18
Answer:
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