## 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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