Cubic Equations
An equation such as \(ax^3=k\) has exactly one real solution, the cube root, so \(2x^3=432\) gives \(x=6\). Cube roots can be negative, unlike square roots. In NSW Year 10 Advanced maths you solve cubic equations exactly and as decimals, and graph cubic curves of the form \(y=a(x-b)^3+c\) (NESA outcome MA5‑NLI‑P‑01).
You will solve equations such as \(x^3=-\dfrac{27}{64}\), \(3(x+2)^3=-24\) and \(6x^3=90\), giving exact cube roots and rounded answers. You will find the equation of a cubic of the form \(y=ax^3+c\) from two points, describe transformations of \(y=x^3\), and find where two cubic curves meet. Problems include the radius of a spherical tank near Tamworth, a box with a square base and a cube whose volume grows. It is part of Year 10 Advanced Non Linear Relationships Functions and Graphs.
Every question with a fully worked solution.
- Cubic Equations - Video - Drawing Cubic Graphs Watch
Theory
A simple cubic equation such as \(3x^3 + 4 = 28\) is solved by making \(x^3\) the subject and taking the cube root; there is always exactly one real solution. This Year 10 Advanced maths page covers cubic equations, the graphs of \(y = ax^3 + c\) and \(y = a(x - b)^3 + c\) with their point of inflection, finding equations from points, and volume problems.
The cube root \(\sqrt[3]{k}\) is the number whose cube is \(k\): \(\sqrt[3]{64} = 4\) and \(\sqrt[3]{-64} = -4\). Unlike a square root, a cube root exists for negative numbers, and every real number has exactly one real cube root.
So \(x^3 = k\) always has exactly one real solution, \(x = \sqrt[3]{k}\). Compare \(x^2 = k\), which has two, one or no real solutions.
The graph of \(y = x^3\) rises from the bottom left to the top right and flattens momentarily at the origin, where the curve changes the way it bends. That point is the point of inflection.
\(y = a(x - b)^3 + c\) is \(y = x^3\) dilated by \(a\) (and reflected in the \(x\)-axis if \(a \lt 0\)), then moved \(b\) right and \(c\) up. Its point of inflection is \((b,\ c)\).
It comes up if you are heading for Advanced or Extension maths in Years 11 and 12.
Solving the basic forms:
Volume formulas that lead to cubic equations:
How to solve a cubic equation of this type
- Undo the adding or subtracting, then the multiplying or dividing, until a cube stands alone: \(x^3 = k\) or \((\ldots)^3 = k\).
- Take the cube root of both sides (it keeps the sign of \(k\)).
- Finish solving the linear equation that is left, and give the answer exactly or rounded as asked.
Problem types
- Sketching \(y = a(x - b)^3 + c\): plot the point of inflection \((b,\ c)\); decide rising (\(a \gt 0\)) or falling (\(a \lt 0\)); find the \(y\)-intercept (\(x = 0\)) and the single \(x\)-intercept (\(y = 0\)). A table of values helps over a given domain.
- Matching equations to features: check the point of inflection and whether the curve rises or falls.
- Finding \(a\) and \(c\): substitute two points and solve simultaneously (Example 3). If one point is the point of inflection \((b,\ c)\), it gives \(c\) straight away.
- Intersections: a horizontal line \(y = d\) meets a cubic of this type once; set the expressions equal and solve. Two cubics such as \(y = px^3 + q\) and \(y = rx^3 + s\): collect the \(x^3\) terms, solve for \(x^3\), then \(x\).
- Volumes: substitute into the volume formula, isolate the cube, take the cube root. If the volume is scaled by \(k\), each length is scaled by \(\sqrt[3]{k}\).
- Expanding \((s + d)^3\): \((s + d)^3 = s^3 + 3s^2 d + 3sd^2 + d^3\). A change in a cube's volume gives a quadratic in \(s\) after the \(s^3\) terms cancel.
(i) \(3x^3 + 7 = -17\)
(ii) \((2x - 1)^3 = 125\)
(iii) \(5x^3 = 60\), exactly and correct to \(2\) decimal places.
Make the cube the subject, then take the cube root of both sides. A cube root of a negative number is negative.
(i)
| \(3x^3\) | \(=\) | \(-24\) |
| \(x^3\) | \(=\) | \(-8\) |
| \(x\) | \(=\) | \(\sqrt[3]{-8} = \mathbf{-2}\) |
(ii) The bracket is the thing being cubed:
| \(2x - 1\) | \(=\) | \(\sqrt[3]{125} = 5\) |
| \(2x\) | \(=\) | \(6\) |
| \(x\) | \(=\) | \(\mathbf{3}\) |
(iii)
| \(x^3\) | \(=\) | \(12\) |
| \(x\) | \(=\) | \(\sqrt[3]{12}\) |
| \(\) | \(\approx\) | \(\mathbf{2.29}\) |
Each equation has exactly one real solution: every real number has exactly one real cube root.
Reflect \(y = x^3\) in the \(x\)-axis, then translate \(1\) unit left and \(8\) units up. The point of inflection moves to \((-1,\ 8)\), and the curve falls from left to right because \(a = -1 \lt 0\).
| \(y\text{-intercept: } y\) | \(=\) | \(-(0 + 1)^3 + 8 = 7\) |
| \(x\text{-intercept: } (x + 1)^3\) | \(=\) | \(8\) |
| \(x + 1\) | \(=\) | \(2\) |
| \(x\) | \(=\) | \(1\) |
Substitute both points:
| \(a + c\) | \(=\) | \(-1\quad (1)\) |
| \(-8a + c\) | \(=\) | \(26\quad (2)\) |
Subtract (2) from (1) to remove \(c\):
| \(9a\) | \(=\) | \(-27\) |
| \(a\) | \(=\) | \(-3\) |
| \(c\) | \(=\) | \(-1 - (-3) = 2\) |
So \(y = \boldsymbol{-3x^3 + 2}\). For the \(x\)-intercept:
| \(3x^3\) | \(=\) | \(2\) |
| \(x\) | \(=\) | \(\sqrt[3]{\dfrac{2}{3}} \approx \mathbf{0.87}\) |
The height is \(2r\), so \(V = \pi r^2 h = \pi r^2 (2r) = 2\pi r^3\).
| \(2\pi r^3\) | \(=\) | \(1000\) |
| \(r^3\) | \(=\) | \(\dfrac{500}{\pi}\) |
| \(r\) | \(=\) | \(\sqrt[3]{\dfrac{500}{\pi}}\) |
| \(\) | \(\approx\) | \(\mathbf{5.42}\text{ cm}\) |
Check: \(2\pi(5.42)^3 \approx 1000\). Keep \(\pi\) in the calculator until the end.
Common pitfalls
Frequently asked questions
How do you solve a cubic equation like 2x^3 = 54?
Divide both sides by 2 to get x cubed equals 27, then take the cube root of both sides: x = 3. There is only one real solution.
Can you take the cube root of a negative number?
Yes. A negative number cubed is negative, so the cube root of a negative number is negative. For example, the cube root of negative 8 is negative 2.
What is a point of inflection on a cubic graph?
It is the point where the curve changes the way it bends, from bending one way to bending the other. For y = a(x - b) cubed + c it is the point (b, c).
Why does x^3 = k have only one solution?
Cubing keeps the sign: positive numbers have positive cubes and negative numbers have negative cubes, and larger numbers have larger cubes. So each value of k is the cube of exactly one real number.
How do you find the radius of a sphere from its volume?
Substitute the volume into V = four thirds pi r cubed, make r cubed the subject by multiplying by 3 and dividing by 4 pi, then take the cube root.