Question

Let r, s and t be the three roots of the equation 8x³+ 1001x+2008=0 . Find the value of 1001-(r+s)³-(s+t)³-(t+r)³​

Answer 1

Step-by-step explanation:

Given :-

r, s and t be the three roots of the equation 8x³+ 1001x+2008=0 .

To find :-

Find the value of 1001-(r+s)³-(s+t)³-(t+r)³?

Solution :-

Given Cubic equation is 8x³+1001x+2008=0

On Comparing this with the standard Cubic equation ax³+bx²+cx+d = 0

We have,

a = 8

b = 0

c = 1001

d = 2008

Given roots of the equation are r,s,t

We know that

Sum of the roots = -b/a

=> r+s+t = -0/8

=> r+s+t = 0------------(1)

Sum of the product of two roots taken at a time = c/a

=> rs + st + tr = 1001/8 -------(2)

Product of the roots = -d/a

=> r×s×t = -2008/8

=> rst = -251 ----------(3)

On taking (1)

r+s+t = 0

=> r+s = -t ---------(4)

or

=> s+t = -r ---------(5)

or

=> t+r = -s ---------(6)

Now,

The value of 1001-(r+s)³-(s+t)³-(t+r)³

=> 1001-[(r+s)³+ (s+t)³+ (t+r)³]

from (4),(5)&(6)

=> 1001 -[ (-t)³+(-r)³+(-s)³]

=> 1001-[-t³-r³-s³]

=> 1001-[-(t³+r³+s³)]

=> 1001+(t³+r³+s³)-------(7)

We know that

If a+b+c = 0 then a³+b³+c³ = 3abc

We have r+s+t = 0 then r³+s³+t³ = 3rst

On applying this in (7)

=> 1001+3rst

=> 1001+3(-251) (from (3))

=> 1001+(-753)

=> 1001-753

=> 248

Answer:-

The value of 1001-(r+s)³-(s+t)³-(t+r)³ is 248

Used formulae:-

• The standard Cubic equation is ax³+bx²+cx+d = 0

• Sum of the roots = -b/a

• Sum of the product of the two roots taken at a time = c/a

• Product of the roots = -d/a

• If a+b+c=0 then a³+b³+c³ = 3abc

• (-)quantity×(-)quantity=(+)quantity