The required root will be ax + bx2 + cx + d, provided that the given expression is equal to (ax + bx + cx + d)2, that is equal to

.4

a2x2+2abx+(2ac + b2) x2 + 2 (ad + bc) x3

+(2bd+c2) x2+2cdx + d2.

Hence, equating the coefficients of corresponding powers of x in the last expression and in the expression whose root is required, we have

a2 = 1; 2ab=-4; 2ac + b2=6; 2ad + 2bc=-8;

2bd+c2=9; 2cd-4; d2 = 4.

The first four of these equations are sufficient to determine the values of a, b, c, d; these values are (taking only the positive value of a), a = 1, b = 2, c = 1, d= −2.

The last three equations will be satisfied by the values of a, b, c, d found from the first four, provided the given expression is a perfect square, which is really the case.

Thus the required square root is x3 − 2x2 + x − 2.

198. When any number of terms of a square root have been obtained as many more can be found by ordinary division.

For suppose the expression whose square root is to be found is the square of

The coefficients a1, a,,... a,, can be found by equating the coefficients of the first 2r powers of x in the square of the above to the coefficients of the corresponding powers of x in the given expression.

The square of the above expression is

(a ̧∞” + α2x2-1 +... + α„x2¬r+1)2 + 2 (α ̧x2 +... + ɑ„x”−r+1)

Now, since the coefficient of the highest power of x in R is a", the highest power of x in the expression within square brackets is 2n-2.

Hence the expression within square brackets will not affect any of the terms from which a, a,... ɑër are determined, for the first 2r terms of the given expression extend from x2n to x2n-2r+1.

It therefore follows that if the square of the sum of the first r terms of the root be subtracted from the given expression, and the remainder be divided by twice the sum of the first r terms, the quotient will give the next r terms of the root.

CUBE ROOT.

199. From the identity

(a + b)3 = a3 + 3a2b+3ab2 + b3,

we see that the cube of a binomial expression has four terms, and that when the cube is arranged according to ascending or descending powers of some letter, the cube roots of its extreme terms are the terms of the original binomial.

Hence the cube root of any perfect cube which has only four terms can be written down by inspection, for we have only to arrange the expression according to powers of some letter and then take the cube roots of its extreme terms.

For example, if 27a6 - 54a5b+36a4b2 - 8a3b3 is a perfect cube its cube root must be 3a2 - 2ab; and by forming the cube of 3a2 - 2ab it is seen that the given expression is really a perfect cube.

When an expression which contains only three different powers of a particular letter is arranged according to powers of that letter, there will be only four terms.

It therefore follows that however many terms there may be in an expression which is a perfect cube, the cube root can be written down by inspection, provided that the expression contains only three different powers of some particular letter.

For example, to find the cube root of

a3+b3+c3+3a2b+3a2c+3ab2+3ac2+6abc +3b2c+3bc2. Arranged according to powers of a, we have

that is,

a3+3a2 (b+c)+3a (b2+c2+2bc) + b3 + c3 + 3b2c+3bc2,
a3+3a2 (b+c)+3a (b+c)2 + (b+c)3.

Hence the required root is a+b+c.

200. To find the cube root of any algebraical expression.

Suppose we have to find the cube root of (A+B)3, where A stands for any number of terms of the root, and B for the rest; the terms in A and B being arranged according to descending (or ascending) powers of some letter, so that every term of A is of higher (or lower) degree in that letter than any term of B.

Also suppose the terms in A are known, and that we have to find the terms in B.

Subtracting A3 from (A + B)3, we have the remainder (3A2 + 3AB + B2) B.

Now from the mode of arrangement it follows that the term of the highest (or lowest) degree in the remainder is 3 x square of the first term of A x first term of B.

Hence to obtain the next term of the required root, that is, to obtain the highest (or lowest) term of B we subtract from the whole expression the cube of that part of the root which is already found and divide the highest (or lowest) term of the remainder by three times the square of the first term of the root.

This gives a method of finding the successive terms of the root after the first; and the first term of the root is clearly the cube root of the first term of the given expression.

For example, to find the cube root of

x6 – 6x3у+21x4y2 − 44x3y3 + 63x2y1 — 54xy3 + 27y6.

The process is written as follows.

x6 – 6x5y + 21x1y2 – 44x3y3 + 63x2y1 — 54xy3 + 27y6 (x2)3 = x6

(x2 – 2xy)3 = x6 – 6x5y +12x1y2 – 8x3y3

(x2 − 2xy + 3y2)3 = x6 − 6x3y + 21x1y2 − 44x3y3 + 63x2y1 − 54xy3 +27y6

Having arranged the given expression according to descending powers of x, we take the cube root of the first term: we thus obtain x2, the first term of the required root.

We then subtract the cube of x2 from the given expression, and divide the first term of the remainder, namely - 6x5y, by 3× (x2)2: we thus obtain - 2xy, the second term of the root.

We then subtract the cube of x2 - 2xy from the given expression, and divide the first term of the remainder by 3 × (x2)2: this will give the third term of the root.

Note. The above rule for finding the cube root of an algebraical expression is rarely, if ever, necessary.

In actual practice cube roots are found as follows.

Take the case just considered; the first and last terms of the root are x and 3y2, the cube roots of the first and last terms of the given expression; also the second term of the root will be found by dividing the second term of the given expression by 3 × (2)2, so that the second term of the root is 2xy.

Hence, if the given expression is really a perfect cube, it must be (x2 - 2xy + 3y2)3, and it is easy to verify that (x2 - 2xy + 3y2)33 is equal to the given expression.

Again, to find the cube root of

xo − 6x3y + 15x1y2 — 29x3y3 + 51x*y* − 60x*y* + 64x3y®

63x2y+27xy - 27y3.

If the given expression is really a perfect cube the first and last terms of the root must be / and /-27y° respectively, that is x3 and - 3y3.

The second term of the root must be - 6x3y ÷ 3 (x3)2 2xy; and the term next to the last must be 27xy+3(-3y3)2 = + xy3.

=

Hence the given expression, if a

must be (-2x2y + xy2 - 3y3)3;

cube at all,

and by expanding

(x3 — 2x3y + xy2 — 3y3) it will be found that the given expression is really a perfect cube.

201. From the identity [see Art. 249]

(a + b)" = a" + na"-1 b + terms of lower degree in a,

it is easy to shew, as in Articles 196 and 200, that the nth root of any algebraical expression can be found by the following

Rule. Arrange the expression according to descending or ascending powers of some letter, and take the nth root of the first term: this gives the first term of the root.

Also, having found any number of terms of the root, subtract from the given expression the nth power of that part of the root which is already found, and divide the first term of the remainder by n times the (n-1)th power of the first term of the root: this gives the next term of the root.

EXAMPLES XIX.

Write down the square roots of the following expressions:

4. 25a +961 + 4c2 + 12b2c2 - 20c2a2 - 30a2b2.