Question

How to solve qubic equation:

r^3 +3r^2 + 3r + 1 = 0

The calculator shows answer to be -1, but i need to know the steps, please solve with steps.

Answer 1

ok

Answer 2

r(r^2+3r+3)+1=o

Answer 3

do u know the x method

Answer 4

co the durand-kerner method

Answer 5

this problem is difficult

Answer 6

can't you just use synthetic division to show that -1 is a zero?

Answer 7

that is what I would do

Answer 8

also then I am left with a quadratic which is easily solvable.

Answer 9

i think you can write as
\[\Large (r+1)^3=0\]

Answer 10

which means
\[\Large (r+1)(r+1)(r+1)=0\]

Answer 11

now
\[\Large r+1=0\]
\[\Large r+1=0\]
\[\Large r+1=0\]

solving
\[\Large r=-1\]

so you have three same roots
\[\Large r=-1\]
\[\Large r=-1\]
\[\Large r=-1\]

Answer 12

sami-21 is correct

Answer 13

my approach would also work

Answer 14

note i used the following
\[\Large (a+b)^3=a^3+b^3+3a^2b+3ab^2\]
to write
\[\Large (r^3+3r^2+3r+1)=(r+1)^3\]
i hope it is clear

Answer 15

@onaogh did you get it

Answer 16

@sami-21, suppose we didn't have calculators, and you didn't knew that the roots were -1, how would u solve it then ??

Answer 17

@onaogh i did solve it without any calculator !

Answer 18

even if you want to know
then the one factor must be factor of the constant term
here the constant term is +1 its factor can be +1 or -1
so use both of them in them equation if by any value it gets zero it means it is root.

Answer 19

use the p/q possible rationals thm
that would lead you to +1 or -1

Answer 20

can you try this:
solve this: r^3 +19r^2 + 6r + 7 = 0

Answer 21

yes sure

Answer 22

ok in this case old methods wouldn't work because of imaginary roots .

Answer 23

i hope this was not in your book ??

Answer 24

I would test 7, -7, 1, -1 as possible factors

Answer 25

Suppose I have 4 unique integer roots, what is the strategy to guess the roots, I thought Vieta and Rational Roots would help?

Answer 26

like what i said do duran-kerner method

Answer 27

no, i made up that equation, i want to see how to be able to solve with any coefficients.
@CJVelasco, yeah googling around and found few methods to solve.

Answer 28

the answers would be in decimal

Answer 29

for 4 your equation has to be a fourth power

Answer 30

i knew that :) thats why i asked it cannot be from book :)

Answer 31

thanks guys for your time.
@sami-21 , thanks a lot

Answer 32

btw imaginary numbers are valid solutions aka as complex solutions
traditional methods would reveal them

Answer 33

thanks @precal you are right ! traditional methods works for cube roots ! and can be found easily .but in the question given above it was not easy .i wrote the sentence in the context of the given question.
Newton method is very fast and good method to approximate the rots of higher order polynomials.!