Question

If a +b+c=6, a² + 62 + c = 34,
then ab + bc + ca =​

Answer 1

Answer:

If a+b+c=6 and a²+b²+c²=14 and a³+b³+c³=36, what is the answer of abc?

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I love modest problems like this. On the face of it the question requires nothing more than a little clever algebra, and indeed that is all that is needed, but we can make it a little more interesting by generalising the problem. First though: A note on notation.

Notation

As it turns out the power sums are well known enough that they have their own symbol

pn(x)=∑k=1Nxnk,(1)

these are called power sum symmetric polynomials.

Within the purview of symmetric polynomials are another type called the elementary symmetric polynomial, written

[math]\displaystyle e_n(x)=\sum_{\begin[/math]

If a+b+c=4, a²+b²+c²=6 and a³+b³+c³=8 then find [a⁴+b⁴+c⁴] where [.] denotes gif?

If a+b+c=1 , a²+b²+c²=2 , a³+b³+c³=3 then what is the value of a⁴+b⁴+c⁴?

If a+b+c=6, a²+b²+c²=26 and a³+b³+c³=90, then what is the value of a,b and c?

If a³+b³=10 and a²+b²=7, then is a+b=?

If a+b+c = 12 and a^2 + b^2 + c^2 = 64, find the value of ab+bc+AC?

Answer: 6

Solution:

In an earlier answer (May 6, 2018) to a question by the same author, I derived the following two relations:

a³+b³+c³ -3abc = (a+b+c)³ -3(ab+bc+ca) (a+b+c)………………………………..…...(1)

(a+b+c)² = a² + b² + c² + 2 (ab+bc+ca)…………………………………………………….(2)

Using the above two equations (1) and (2), we can find the answer to abc. We are given,

a+b+c=6, a²+b²+c²=14 and a³+b³+c³=36……………………………..…………………(A)

From eq.(2),

2 (ab+bc+ca) = (a+b+c)² - (a² + b² + c²)

= 6² - 14 = 36–14=22 [Using eq.(A)]

Or, ab+bc+ca = 11………………………………………………………………………………(B)

We now substitute the numerical values given in equations (A) a

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Given that( a+b+c)=6, a^2+b^2+c^2 =14, a^3+b^3+c^3 =36

There is an identity

a^3+b^3+c^3 -3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) ………(1)

Taking (a+b+c) =6 ,squarreing both sides

(a+b+c)^2 = 36

a^2+b^2+c^2+2ab+2bc+2ca =36

14 +2ab+2bc+2ca=36

2(ab +bc+ca)= 36–14=22

ab+bc+ ca= 22/2

ab+bc+ca= 11,………..(2)

Putting the value of ab+bc+ ca =11 ,a^2+b^2+c^2 , a+b+c,a^3+b^3+c^3 in eq(1)

36–3abc=6(14–11)

36–3abc=6×3

-3abc =18–36

-3abc=-18

abc= 18/3=6.