If a+b+c=6,ab+bc+ca=11,then what is a3+b3+c3-3abc

Work Shown:

a+b+c = 6

(a+b+c)^2 = 6^2

(a+b+c)(a+b+c) = 36

a(a+b+c)+b(a+b+c)+c(a+b+c) = 36

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

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

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

a^2+b^2+c^2+2*11 = 36

a^2+b^2+c^2+22 = 36

a^2+b^2+c^2 = 36-22

a^2+b^2+c^2 = 14

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

a^3+b^3+c^3-3abc = 6*(14 - 11)

a^3+b^3+c^3-3abc = 18

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

~

b

step-by-step explanation:

it have 5 sides and 5 identicle angle.the rest does not have the same property as b.

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