Question

If mm, nn and pp are the roots of the polynomial x^3+5x-8x
3
+5x−8 and sum of the roots is 00, then find the value of m^3+n^3+p^3.m
3
+n
3
+p
3
.

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{}$

$\mathsf{m,n\;and\;p\;are\;the\;roots\;of\;x^3+5x-8}$

$\underline{\textbf{To find:}}$

$\textsf{The value of}$

$\mathsf{m^3+n^3+p^3}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,\;x^3+5x-8}$

$\mathsf{Sum\;of\;the\;roots=\dfrac{-(0)}{1}}$

$\implies\mathsf{m+n+p=0}$

$\mathsf{Product\;of\;the\;roots=\dfrac{-(-8)}{1}}$

$\implies\mathsf{mnp=8}$

$\textsf{We know that,}$

$\boxed{\mathsf{a+b+c=0\;\implies\;a^3+b^3+c^3=3\,abc}}$

$\implies\mathsf{m^3+n^3+p^3}$

$\mathsf{=3\;mnp}$

$\mathsf{=3(8)}$

$\mathsf{=24}$

$\implies\boxed{\mathsf{m^3+n^3+p^3=24}}$

Answer 2

Step-by-step explanation:

Given : x³+5x-8

To find: If m, n and p are the roots of the polynomial and sum of the roots is 0, then find the value of m³+n³+p³.

Solution:

We know that there is a relation between zeros of polynomial with coefficient of polynomial.

On comparison with standard cubic polynomial ax³+bx²+cx+d

a=1

b=0

c=5

d=-8

ATQ

m, n and p are roots of polynomial,

Step 1: Write the relationship of zeros with coefficient of cubic polynomial.

$\bold{m + n + p = - \frac{b}{a} = 0}...eq1$

and

$\bold{mn + np + mp = \frac{c}{a} = 5 }...eq2$

and

$\bold{mnp = \frac{ - d}{a} = 8}...eq3$

Step 2: Take cube of eq1 both sides

$( {m + n + p)}^{3} = {0}^{3}$

open the identity

$\bold{\red{( {a + b)}^{3} = {a}^{3} + 3 {a}^{2} b + 3a {b}^{2} + {b}^{3}}}$

let (m+n) =a and p=b

${(m + n)}^{3} + 3 {(m + n)}^{2} p + 3(m + n) {p}^{2} + {p}^{3} = 0$

again apply the same identity on (m+n)³ and (m+n)²

${m}^{3} + 3 {m}^{2}n + 3m {n}^{2} + {n}^{3} + 3 {(m + n)}^{2} p + 3(m + n) {p}^{2} + {p}^{3} = 0$

or

${m}^{3} + {n}^{3} + {p}^{3} + 3 {m}^{2} n+ 3m {n}^{2} + 3 ( {m}^{2} + 2mn + {n}^{2}) p + 3m {p}^{2} +3 n {p}^{2} = 0$

or

${m}^{3} + {n}^{3} + {p}^{3} + 3 {m}^{2} n+ 3m {n}^{2} + 3 {m}^{2}p + 6mnp + 3{n}^{2} p + 3m {p}^{2} +3 n {p}^{2} = 0$

Add and subtract 3mnp and manipulate the terms so that value of eq2 can be used

${m}^{3} + {n}^{3} + {p}^{3} + 3m( {m} n + {m}p + np )+ 3n(mp + {n} p + 3m {n})+3p( n {p}+m {p}+mn)-3mnp= 0$

put the value of eq2 and eq3

${m}^{3} + {n}^{3} + {p}^{3} + 3m(5 )+ 3n(5)+3p(5)-3(8)= 0$

or

${m}^{3} + {n}^{3} + {p}^{3} + 15m+ 15n+15p-24= 0$

${m}^{3} + {n}^{3} + {p}^{3} + 15(m+ n+p)-24= 0$

put the value of eq1

${m}^{3} + {n}^{3} + {p}^{3} + 15(0)-24= 0$

or

${m}^{3} + {n}^{3} + {p}^{3} =24$

Final answer:

$\bold{\red{{m}^{3} + {n}^{3} + {p}^{3} =24} }$

Hope it helps you.

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