Let k are roots of equation 8x3+1001x+2008=0. Then value of (r + s)3+(s + t)3+(t + r)3 is 7k3 (where k is at ten's place). Then k=

Question

Let k are roots of equation 8x3+1001x+2008=0. Then value of (r + s)3+(s + t)3+(t + r)3 is 7k3 (where k is at ten's place). Then k= (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

r+s+t=0, rst=(- 2008/8)=-251 Now Let r+s=x s+t=y t+r=z ∴ x+y+z=2(r + s + t)=0 ∴ x3+y3+z3=3xyz =3(r + s) (s + t) (t + r) =3(- t) (- r) (- s)=-3 rst =-3× (- 251)=753

Q. Let $k$ are roots of equation $8 x^{3} + 1001 x + 2008 = 0.$ Then value of $(r + s)^{3} + (s + t)^{3} + (t + r)^{3}$ is $7 k 3$ (where k is at ten's place). Then $k =$

$r + s + t = 0, rs t = \frac{- 2008}{8} = - 251$
Now Let
$r + s = x$
$s + t = y$
$t + r = z$
$∴ x + y + z = 2 (r + s + t) = 0$
$∴ x^{3} + y^{3} + z^{3} = 3 x yz$
$= 3 (r + s) (s + t) (t + r)$
$= 3 (- t) (- r) (- s) = - 3 rs t$
$= - 3 \times (- 251) = 753$

Q. Let k are roots of equation 8x3+1001x+2008=0. Then value of (r+s)3+(s+t)3+(t+r)3 is 7k3 (where k is at ten's place). Then k=

r+s+t=0,rst=8−2008​=−251 Now Let r+s=x s+t=y t+r=z ∴x+y+z=2(r+s+t)=0 ∴x3+y3+z3=3xyz =3(r+s)(s+t)(t+r) =3(−t)(−r)(−s)=−3rst =−3×(−251)=753

Q. Let $k$ are roots of equation $8 x^{3} + 1001 x + 2008 = 0.$ Then value of $(r + s)^{3} + (s + t)^{3} + (t + r)^{3}$ is $7 k 3$ (where k is at ten's place). Then $k =$

$r + s + t = 0, rs t = \frac{- 2008}{8} = - 251$
Now Let
$r + s = x$
$s + t = y$
$t + r = z$
$∴ x + y + z = 2 (r + s + t) = 0$
$∴ x^{3} + y^{3} + z^{3} = 3 x yz$
$= 3 (r + s) (s + t) (t + r)$
$= 3 (- t) (- r) (- s) = - 3 rs t$
$= - 3 \times (- 251) = 753$