Lesson Video: Solving Cubic Equations: Taking Cube Roots | Nagwa Lesson Video: Solving Cubic Equations: Taking Cube Roots | Nagwa

# Lesson Video: Solving Cubic Equations: Taking Cube Roots Mathematics • 8th Grade

In this video, we will learn how to solve cubic equations using the cube root property.

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### Video Transcript

In this video, we will learn how to solve cubic equations using the cube root property. We begin by recalling that the cube root of a number π written as shown is the number whose cube is π. In other words, the cube root of π cubed is equal to π. We can use this to simplify or evaluate expressions. For example, we know that two cubed is equal to eight, which means that the cube root of eight is two.

This is not the only use of the cube root, as we can also use this idea to solve equations. For example, imagine we are told that the volume of a cube is 64 centimeters cubed. We can then say that the cube has side lengths π₯ centimeters, giving us the equation π₯ cubed is equal to 64. Cube rooting both sides of this equation, we know that the cube root of π₯ cubed is π₯. And the cube root of 64 is four. We can therefore conclude that the cube has side lengths of four centimeters.

At this stage, it is worth noting that the equation π₯ cubed equals π will only have one solution for any real value of π. This is because cubing a positive number gives a positive answer, and cubing a negative number gives a negative answer. This is different to solving the equation π₯ squared equals π as this has two solutions when π is a positive real number. For example, when π₯ squared is equal to nine, we know that π₯ is equal to positive or negative three, since three squared is equal to nine and negative three squared is equal to nine. Letβs now look at an example where we need to solve a cubic equation.

Solve the cubic equation π₯ cubed equals eight for all rational numbers.

We recall that a rational number is a number that can be expressed as the quotient or fraction of two integers. In this question, we need to solve the equation π₯ cubed equals eight and we will do so by firstly taking the cube root of both sides. We know that the cube root of π₯ cubed is simply equal to π₯. And since two cubed is equal to eight, the cube root of eight is equal to two. The solution to the cubic equation π₯ cubed is equal to eight is therefore π₯ is equal to two.

In our next example, we will see how to solve a cubic equation where the variable cubed is a fraction.

Solve π₯ cubed is equal to 27 over eight.

In order to solve this equation, we will begin by taking the cube root of both sides. We recall that the cube root of π₯ cubed is π₯. So π₯ is equal to the cube root of 27 over eight. Next, we recall that the cube root of π over π, where π is nonzero is equal to the cube root of π over the cube root of π. We know that since two cubed is equal to eight, the cube root of eight is two. And since three cubed is 27, the cube root of 27 is three. Since our equation simplifies to π₯ is equal to the cube root of 27 over the cube root of eight, then π₯ is equal to three over two or three-halves. This is the single solution to the equation π₯ cubed is equal to 27 over eight.

In our next example, we will solve a cubic equation by first rearranging.

Given that π₯ exists in the set of real numbers and negative π₯ over 10 is equal to 100 over π₯ squared, determine the value of π₯.

In order to solve the given equation, weβll begin by cross multiplying. This is the same as multiplying both sides of the equation by 10π₯ squared. On the left-hand side, we have negative π₯ multiplied by π₯ squared. And on the right-hand side, 100 multiplied by 10. This simplifies to negative π₯ cubed is equal to 1000. Multiplying through by negative one so that our π₯-term is positive, we have π₯ cubed is equal to negative 1000. We can then take the cube root of both sides of this equation. The cube root of π₯ cubed is π₯. And noting that negative 10 cubed is equal to negative 1000, then the cube root of negative 1000 is negative 10. If negative π₯ over 10 is equal to 100 over π₯ squared, then the value of π₯ is negative 10.

We can check this answer by substituting our value of π₯ back in to the original equation.

So far in this video, we have only doubt with simple equations involving cubes. However, it is possible for operations to occur inside the cubic operation. In general, we can solve equations of the form ππ₯ plus π all cubed plus π is equal to π provided that π is not equal to zero and we can find the cube root of π minus π. We solve an equation of this type using the following four steps.

Firstly, we subtract π from both sides, giving us ππ₯ plus π all cubed is equal to π minus π. Secondly, we take the cube root of both sides of the equation to get ππ₯ plus π is equal to the cube root of π minus π. Next, we subtract π from both sides such that ππ₯ is equal to the cube root of π minus π minus π. And finally, we divide through by π so that π₯ is equal to the cube root of π minus π minus π all divided by π. Letβs now consider how we can apply this in practice.

Find the value of π¦ given that two π¦ minus 14 all cubed minus 36 is equal to 28.

The equation in this question is written in the form ππ₯ plus π all cubed plus π is equal to π. We know that we can solve equations of this type by rearranging to make π₯ the subject. In this question, the variable is π¦. So we will follow a similar method in order to make π¦ the subject. We begin by adding 36 to both sides of our equation. This gives us two π¦ minus 14 all cubed is equal to 28 plus 36. The right-hand side simplifies to 64, and we can now take the cube root of both sides of this equation. On the left-hand side, we have two π¦ minus 14. And since four cubed is equal to 64, the cube root of 64 is four. Our equation simplifies to two π¦ minus 14 equals four.

Next, we add 14 to both sides. Two π¦ is equal to four plus 14, which is equal to 18. Finally, we can divide through by two such that π¦ is equal to nine. If two π¦ minus 14 all cubed minus 36 is equal to 28, then the value of π¦ is nine. We can check this answer by substituting π¦ is equal to nine back into our original equation. When we do this, we have two multiplied by nine minus 14 inside our parentheses. This is equal to four, and four cubed minus 36 is indeed equal to 28. This confirms that the solution to the equation is π¦ is equal to nine.

In our final example, weβll solve a similar problem. However, this time the coefficient of the variable is negative.

Find the value of π₯ given that 15 minus three π₯ all cubed plus two is equal to 29.

In order to answer this question, we will first rearrange the equation so that the cubed term is isolated on the left-hand side. We do this by subtracting two from both sides, giving us 15 minus three π₯ all cubed is equal to 29 minus two. 29 minus two is equal to 27. And we can now take the cube roots of both sides of the equation. The cube root of 15 minus three π₯ all cubed is simply 15 minus three π₯. And since three cubed is 27, the cube root of 27 is equal to three. Our equation simplifies to 15 minus three π₯ is equal to three.

In order to solve for π₯, we can now subtract 15 from both sides so that negative three π₯ is equal to three minus 15. This in turn simplifies to negative three π₯ is equal to negative 12. And dividing through by three, we have π₯ is equal to four. The value of π₯ that satisfies the equation 15 minus three π₯ all cubed plus two is equal to 29 is four.

We will now finish this video by recapping the key points. We saw in this video that we can solve equations by taking the cube roots of both sides of the equation. In particular, if π₯ cubed is equal to π, then π₯ is equal to the cube root of π. We also saw that unlike the square root, taking cube roots of both sides of an equation gives a unique solution. Finally, to solve a cubic equation of the form ππ₯ plus π all cubed plus π equals π, where π, π, π, and π are constants and π is not equal to zero, we rearrange the equation to isolate π₯. This gives us π₯ is equal to the cube root of π minus π minus π all divided by π.

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