Question

if x=1/3+√8 find the value of x cube - 2x square - 7x +5

Answer 1

I hope this answer would be helpful for you

Answer 2

Answer:

Step 1: Give data

If $x= \frac{1}{3} +\sqrt{8}$

Step 2: To find

Find the value for $x^{3}-2x^{2} -7x+5$.

Step 3: Finding solution

$x= \frac{1}{3} +\sqrt{8}$

$x=\frac{1}{3+\sqrt{8} }\times\frac{3-\sqrt{8} }{3-\sqrt{8} }$

x $=\frac{3-\sqrt{8} }{9-8}$  

x = 3 $-\sqrt{8}$

Substitute the value of x in the given equation to find the answer for the given question.

Now, $x^{3}-2x^{2} -7x+5$ =  $({} 3-\sqrt{8} )^{3} -(3-\sqrt{8}) ^{2}-7(3-\sqrt{8} )+5$    

we know that,

$(a-b)^{3}=a^{3} -b^{3} -3a^{2} b+3ab^{2}$

$(a-b)^{2} =a^{2} +b^{2} -2ab$

Here a =3 and b =$\sqrt{8}$  in the above formula

Those two formula are used to express the above equation.

Now we get,

$({} 3-\sqrt{8} )^{3}$  = $(3)^{3} -(\sqrt{8}) ^{3} -3(3)^{2} (\sqrt{8} )+3(3)(\sqrt{8}) ^{2})$

               = 27 - 8$\sqrt{8}$ - 27$\sqrt{8}$ + 9 (8)

                 27 - 8$\sqrt{8}$ - 27$\sqrt{8}$ + 72

$(3-\sqrt{8}) ^{2}$ = $3^{2} +(\sqrt{8}) ^{2} -2(3)(\sqrt{8} )$

               = 9 + 8 - 6$\sqrt{8}$

               =17 - 6$\sqrt{8}$

substitute  those expression value in the equation to find the value.

$x^{3}-2x^{2} -7x+5$ =  $({} 3-\sqrt{8} )^{3} -(3-\sqrt{8}) ^{2}-7(3-\sqrt{8} )+5$    

                            =   27 - 8$\sqrt{8}$ - 27$\sqrt{8}$ + 72 - 2( 17 - 6$\sqrt{8}$ ) - 7 (3 -$\sqrt{8}$ ) + 5

                           = 27 -8$\sqrt{8}$  -27$\sqrt{8}$ + 72 - 34 + 12$\sqrt{8}$  -21 +7$\sqrt{8}$ +5

                           = 104 - 55 + 19$\sqrt{8}$  -35$\sqrt{8}$

                           = 49 -16$\sqrt{8}$

Hence the value for the given equation is  $49-16\sqrt{8}$.

To learn more...

1) If x=1/2-sq root 3,find the value of x3-2x2-7x+5

https://brainly.in/question/100089

2) Prove that x³-2x²-7x+5=3 when x=1/(2-√3)

https://brainly.in/question/125173