Miscellaneous Examples of Algebra: Part 3 for IITJEE Mains

Many identities can be readily established by making use of  the properties of the cube roots of unity; as usual these will be denoted by 1, \omega, \omega^{2}.

Problem 1:

Show that (x+y)^{7} - x^{7} - y^{7} = 7xy(x+y)(x^{2}+xy+y^{2})^{2}

Solution 1:

The expression, E, on the left vanishes when x=0, y=0, x+y=0; hence, it must contain xy(x+y) as a factor.

Putting  x = {\omega}y, we have

E=((1+\omega)^{7}-\omega^{7}-1)y^{7} = ((-\omega^{2})^{7}-\omega^{7}-1)y^{7}=(-\omega^{2}-\omega-1)y^{7}

Hence, E contains x - {\omega}y as a factor; and similarly, we may show that it contains x - \omega^{2}y as a factor; that is, E is divisible by

(x-{\omega}y)(x-\omega^{2}y), or (x^{2}+xy+y^{2}).

Further, E being of seven, and xy(x+y)(x^{2}+xy+y^{2}) of five dimensions, the remaining factor must be of the form A(x^{2}+y^{2})+Bxy, thus,

(x+y)^{7}-x^{7}-y^{7} = xy(x+y)(x^{2}+xy+y^{2})(Ax^{2}+Bxy+Ay^{2}).

Putting x=1, y=1, we have 21 = 2A + B; putting x = 2, y = -1, we have 21= 5A-2B, and hence, A=7, B=7;

(x+y)^{7} - x^{7} - y^{7} = 7xy(x+y)(x^{2}+xy+y^{2}).

Problem 2:

Show that the product of a^{3} + b^{3} + c^{3} -3abc and x^{3} + y^{3} + z^{3} -3xyz can be put in the form A^{3} + B^{3} + C^{3} -3ABC.

Solution 2:

The product = (a+b+c)(a+{\omega}b + \omega^{2}c)(a + \omega^{2}b + {\omega}c) \times (x+y+z)(x+{\omega}y+\omega^{2}z)(x+\omega^{2}y + {\omega}z)

By taking these six factors in the pairs (a+b+c)(x+y+z),

(a+{\omega}b+\omega^{2}c)(x+\omega^{2}y+{\omega}z),

and (a + \omega^{2}b + {\omega}c)(x + {\omega}y + \omega^{2}z),

we obtain the partial products:

A+B+C, A + {\omega}B + \omega^{2}C, and A + \omega^{2}B + {\omega}C

where  A =ax + by + cz, B = bx + cy + az, and C = cx + ay + bz

Thus, the product  = (A + B + C)(A + {\omega}B + \omega^{2}C)(A + \omega^{2}B + {\omega}C)

which, in turn, equals A^{3} + B^{3} + C^{3} + 3ABC

More esoteric algebraic miscellany is planned for you!

Nalin Pithwa

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