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.!

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