HardMathematicsHigh School (9–10)GeneralAlgebra › Algebraic Identities and FactorisationNumericalConceptualBoard Exam

Show that if x + y + z = 0, then x³ + y³ + z³ = 3xyz. Use it to find 9³ + (-4)³ + (-5)³.

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Show that if $x + y + z = 0$ , then $x^3 + y^3 + z^3 = 3xyz$ . Use it to find $9^3 + (-4)^3 + (-5)^3$ .

Need a reminder? The key ideaShow

$(x \pm y)^2 = x^2 \pm 2xy + y^2$ , $x^2 - y^2 = (x + y)(x - y)$ , $(x \pm y)^3 = x^3 \pm y^3 \pm 3xy(x \pm y)$ , $x^3 \pm y^3 = (x \pm y)(x^2 \mp xy + y^2)$ , and if $x + y + z = 0$ then $x^3 + y^3 + z^3 = 3xyz$ .

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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Show that if x + y + z = 0, then x³ + y³ + z³ = 3xyz. Use it to find 9³ + (-4)³ + (-5)³.

The worked solution

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