If #a+b+c=7#, #a^2+b^2+c^2=35# and #a^3+b^3+c^3=151#, then what is #abc# ?

3 Answers
Mr. Whoever · EZ as pi
Apr 17, 2017

#(abc)=-15#

Explanation:

#a+b+c=7#
#5+3+(-1)=7#

#a^2+b^2+c^2=35#
#5^2+3^2+(-1)^2#
#=25+9+1#
# =35#

#a^3+b^3+c^3=151#
#5^3+3^3+(-1)^3#
#=125+27-1#
#=151#

#(abc)#
#=5×3×-1#
#=-15#

George C.
Apr 17, 2017

#abc = -15#

Explanation:

Given:

#{ (a+b+c=7), (a^2+b^2+c^2=35), (a^3+b^3+c^3=151) :}#

Note that:

#(a+b+c)^3#

#= a^3+b^3+c^3+3ab^2+3a^2b+3bc^2+3b^2c+3ca^2+3ac^2+6abc#

#(a+b+c)(a^2+b^2+c^2)#

#= a^3+b^3+c^3+ab^2+a^2b+bc^2+b^2c+ca^2+c^2a#

So:

#6abc = (a+b+c)^3-3(a+b+c)(a^2+b^2+c^2)+2(a^3+b^3+c^3)#

#color(white)(6abc) = 7^3-3(7)(35)+2(151)#

#color(white)(6abc) = 343-735+302#

#color(white)(6abc) = -90#

Dividing both sides by #6# we find:

#abc = -15#

Ratnaker Mehta
Apr 17, 2017

#-15.#

Explanation:

We will use the following Result :

#a^3+b^3+c^3-3abc#

#=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)...............(star)#

Now, #a+b+c=7 rArr (a+b+c)^2=49.#

# rArr a^2+b^2+c^2+2(ab+bc+ca)=49.#

But, we are given that, #a^2+b^2+c^2=35.#

#:. 35+2(ab+bc+ca)=49, or,#

# ab+bc+ca=(49-35)/2=7...(ast)#

Thus, altogether, we have,

#a^3+b^3+c^3=151, a+b+c=7, a^2+b^2+c^2=35..."(given),"#

#&, by (ast), ab+bc+ca=7.#

Using all these in #(star),# we get,

#151-3abc=(7){35-(7)}=196.#

#:. abc=1/3(151-196)=-45/3=-15,# as deived by,

Respcted George C. Sir.

Enjoy Maths.!

Related questions
See all questions in Linear, Exponential, and Quadratic Models
Impact of this question
4297 views around the world
You can reuse this answer
Creative Commons License