if a+b+c = 11, ab + bc + ca = 3 and abc= -135, then what is the value of a³ + b³+ c³ =
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Step-by-step explanation:
a+b+c = 11,
on squareing both side
(a+b+c)² = 11²
a²+b²+c²+2(ab + bc + ca)= 121
a²+b²+c²+2(3)=121
a²+b²+c²+6=121
a²+b²+c²=115
a³ + b³+ c³-3abc= (a+b+c)(a²+b²+c²-ab-bc-ca)
a³ + b³+ c³= (a+b+c)(a²+b²+c²-ab-bc-ca)+3abc
a³ + b³+ c³= (a+b+c){a²+b²+c²-(ab+bc+ca)}+3abc
a³ + b³+ c³= 11(115-3)+3×-135
a³ + b³+ c³=11×112-405
a³ + b³+ c³=1232-405
a³ + b³+ c³= 827
#666
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