Question

if ab+bc+cd=0 for every negative integer a,b,c,d the value of a×b×c×d=​

Answer 1

Step-by-step explanation:

If ab+bc+ca=0, find the value of (1/a²-bc)+(1/b²-ca)+(1/c²-ab) Get ... (ab+bc+ ca)/(a+b+c) abc = (a+b+c) abc/(a+b+c) abc = 0.

Answer 2

Answer:

Required value is 0.

Step-by-step explanation:

Let abcd= x.....(1)

Given,ab+bc+cd=0......(2)

So,ab=x/cd and bc=x/ad and cd=x/bd

From equation (2)

ab+bc+cd=0

we are putting value of ab,bc and cd,

(x/cd)+(x/ad)+(x/bd)=0

→x(1/cd)+(1/ad)+(1/bd)x

x[1/cd)+(1/ad)+(1/bd)]=0

x=0/[1/cd)+(1/ad)+(1/bd)

So,x=0

Therefore abcd=x=0

This is a problem of Algebra.

Some important formulas of algebra,

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)