158. Let a n

represent any radical of the nth degree. Then we may raise this radical to the mth power, by taking it m times as a factor; thus,

But, by the rule for multiplication, this continued product is equal to amn/bm; whence,

We have then, to raise a radical to any power, the following

RULE.

Raise the co-efficient to the required power for a new co-efficient; after this write the radical sign with its primitive index, placing under it the required power of the quantity under the radical sign in the given expression; the result will be the power required.

4

EXAMPLES.

1. (4a3)2 = √(4a3)2 = 1/16a6 = 2a√ a2 = 2a √a.

2. (33/2u) = 35, 3/(2a)5 = 243 3/32a = 486a 3/4a2.

When the index of the radical is a multiple of the exponent of the power to which it is to be raised, the result can be simplified.

For, √2a

=

2a

√2 (Art. 152): hence, in order to square

2a, we have only to omit the first radical sign, which gives

When the index of the radical is divisible by the exponent of the power to which it is to be raised, perform the division, leaving the quantity under the radical sign unchanged.

Extraction of Roots of Radicals of any Degree.

159. By extracting the mth root of both members of equa tion (1), of the preceding article, we find,

Whence we see, that to extract any root of a radical of any degree, we have the following

RULE.

Extract the required root of the co-efficient for a new co-efficient; after this write the radical sign with its primitive index, under which place the required root of the quantity under the radical sign in the given expression; the result will be the root required.

EXAMPLES.

1. Find the cube root of 84/27.

2. Find the fourth root of 13/256.

Ans. 24/3. Ans. 13√4.

159*. If, however, the required root of the quantity under the radical sign cannot be exactly found, we may proceed in the following manner. If it be required to find the mth root of , the operation may be indicated thus,

m

but

equation,

√d=m√d, whence, by substituting in the previous

Consequently, when we cannot extract the required root of the quantity under the radical sign,

Extract the required root of the co-efficient for a new co-efficient; after this, write the radical sign, with an index equal to the product of its primitive index by the index of the required root, leaving the quantity under the radical sign unchanged.

EXAMPLES.

1.

12

√√√√3c=22/3; and, √√√√5c = √5c.

When the quantity under the radical is a perfect power, of the degree of either of the roots to be extracted, the result can be

Different Roots of the same Power.

160. The rules just demonstrated depend upon the principle, that if two quantities are equal, the like roots of those quantities are also equal.

This principle is true so long as we regard the term root in its general sense, but when the term is used in a restricted sense, it requires some modification. This modification is particularly necessary in operating upon imaginary expressions, which are not roots, strictly speaking, but mere indications of operations which it is impossible to perform. Before pointing out these modifications, it will be shown, that every quantity has more than one cube root, fourth root, &c.

It has already been shown, that every quantity has two square roots, equal, with contrary signs.

1. Let x denote the general expression for the cube root of a, and let p denote the numerical value of this root; we have the equations

x3 = a, and x3 = p3.

The last equation is satisfied by making xp.

Observing that the equation x3 = p3 can be put under the form ☛3 — p3 = 0, and that the expression x3-p3 is divisible by zp, giving the quotient, x2 + px + p2, the above equation can be placed under the form

(x − p) (x2 + px + p2) = 0.

Now, every value of x that will satisfy this equation, will satisfy the first equation. But this equation can be satisfied by supposing

hence, we see, that there are three different algebraic expressions

in which p denotes the arithmetical value of

This equation can be put under the form.

which may be satisfied by making either of the factors equal

And if in the equation 23 + p3 = 0, we make p = — p', it becomes x3 — p'3 = 0, from which we deduce

and consequently, the 6th root of a, admits of six different alge braic expressions. If we make

It may be demonstrated, generally, that there are as many different expressions for the nth root of a quantity as there are units in n. If n is an even number, and the quantity is posi tive, two of the expressions will be real, and equal, with contrary signs; all the rest will be imaginary if the quantity is negative, they will all be imaginary.