Simple Formulae and their Application

In the article Simple Formula and their Applications I we dealt with algebraic formulas in the second degree, i.e., formulas related to perfect squares and the sum and difference of two squares. In this article we will be covering the algebraic formulas in the third degree, i.e., formulas related to perfect cubes and the sum and difference of two cubes.

Formula:

$(a+b)^3=a^3+3a^2b+3ab^2+b^3$

Again, $(a+b)^3=a^3+b^3+3ab(a+b)$ [Taking $3ab$ common from $3a^2b+3ab^2$ ]

Proof:

$(a+b)^3=(a+b)(a+b)^2=(a+b)(a^2+b^2+2ab) \\ =a(a^2+b^2+2ab)+b(a^2+2ab+b^2) \\ = a^3+3a^2b+3ab^2+b^3$

Corollary: $a^3+b^3=[a^3+b^3+3ab(a+b)]-3ab(a+b)$

or, $a^3+b^3=(a+b)^3-3ab(a+b)$

Example: Find the cube of $3a+5b$

Solution:

$(3a+5b)^3=(3a)^3+3(3a)^2(5b)+3(3a)(5b)^2+(5b)^3 \\ =27a^3+3(9a^2)(5b)+3(3a)(25b^2)+125b^3 \\ =27a^3+135a^2b+225ab^2+125b^3$

Example: Find the cube of $p+q+r$

Solution:

$(p+q+r)^3=[(p+q)^+r]^3=(p+q)^3+3(p+q)^2r+3(p+q)r^2+r^3 \\ =(p^3+3p^2q+3pq^2+q^3)+3(p^2+q^2+2pq)r+3(p+q)r^2+r^3 \\ =p^3+q^3+r^3+3pq^2+3p^2r+3pr^2+3q^2r+3qr^2+6pqr$

Example: If $x+\dfrac{1}{x}=p$ , show that $x^3+\dfrac{1}{x^3}=p^3-3p$

Solution:

$\because$ $a^3+b^3=(a+b)^3-3ab(a+b)$

$\therefore$ $x^3+\dfrac{1}{x^3}=( x+\dfrac{1}{x})^3-3.x.\dfrac{1}{x}.( x+\dfrac{1}{x}) \\ =(x+\dfrac{1}{x})^3-3(x+\dfrac{1}{x})=p^3-3p$ [ $\because$ $x+\dfrac{1}{x}=p$ ] (Proved)

Example: Simplify $(x-y)^3+(x+y)^3+3(x-y)^2(x+y)+3(x+y)^2(x-y)$

Solution:

Putting $a$ for $x-y$ and $b$ for $x+y$ we have,

The given expression $=a^3+b^3+3a^b+3ab^2$

$= a^3+3a^2b+3ab^2+b^3 \\ =(a+b)^3=[(x-y)+(x+y)]^3=(2x)^3=8x^3$

Example: Show that $x^3+\dfrac{1}{x^3}=-2$ when $\dfrac{x^2+1}{x}=1$

Solution:

$\dfrac{x^2+1}{x}=\dfrac{x^2}{x}+\dfrac{1}{x}=x+\dfrac{1}{x}$

$\therefore$ from the given condition we have,

$x+\dfrac{1}{x}=1 \\ or, \: x^3+\dfrac{1}{x^3}=(x+\dfrac{1}{x})^3-3.x.\dfrac{!}{x}.(x+\dfrac{1}{x}) \\ or, \: 1^3-3.1=1-3=-2$ (Proved)

Formula:

$(a-b)^3= a^3-3a^2b+3ab^2-b^3$

Again, $(a-b)^3=a^3-b^3-3ab(a-b)$ [Taking $-3ab$ common from $-3a^2b+3ab^2$ ]

Proof:

$(a-b)^3=(a-b)(a-b)^2=(a-b)(a^2-2ab+b^2) \\ =a(a^2-2ab+b^2)-b(a^2-2ab+b^2) \\ =a^3-3a^2b+3ab^2-b^3$

Corollary: $a^3-b^3=[a^3-b^3-3ab(a-b)]+3ab(a-b)$

or, $a^3-b^3=(a-b)^3+3ab(a-b)$

Example: Find the cube of $3x-4y$

Solution:

$(3x-4y)^3=(3x)^3-3(3x)^2(4y)+3(3x)(4y)^2-(4y)^3 \\ =27x^3-3(9x^2)(4y)+3(3x)(16y^2)-64y^3 \\ =27x^3-108x^2y+144xy^2-64y^3$

Example: Find the cube of $(a-b-c)$

Solution:

$(a-b-c)^3=[(a-b)-c]^3=(a-b)^3-3(a-b)^2c+3(a-b)c^2-c^3 \\ =(a^3-3a^2b+3ab^2-b^3)-3(a^2+b^2-2ab)c+3(a-b)c^2-c^3 \\ =a^3-b^3-c^3-3a^2b+3ab^2-3a^2c+3ac^2-3b^2c-3bc^2+6abc$

Example: Find the cube of 297

Solution:

$297^3=(300-3)^3=300^3-3^3-3.300.3.(300-3)$ [ $\because$ $(a-b)^3=a^3-b^3-3ab(a-b)$ ] $\\ =27000000-27-810000+8100 \\ =(27000000+8100)-(27+810000) \\ =27008100-810027$

Example: If $\dfrac{x^2-1}{x}=4$ , find the value of $\dfrac{x^6-1}{x^3}$

Solution:

$\dfrac{x^2-1}{x}=\dfrac{x^2}{x}-\dfrac{1}{x}=x-\dfrac{1}{x}$

$\therefore$ $\dfrac{x^2-1}{x}=4$ …(1)

[ $\because$ $\dfrac{x^2-1}{x}=4$ ]

$\dfrac{x^6-1}{x^3}=\dfrac{x^6}{x^3}-\dfrac{1}{x^3}=x^3-\dfrac{1}{x^3} \\ =(x-\dfrac{1}{x})^3+3.x.\dfrac{1}{x}.(x-\dfrac{1}{x})$

$=4^3+3.4=64+12=76.$ [Using (1)]

Exercise 1:

1. Find the cube of:

a) $\dfrac{2}{3a}+\dfrac{3}{5b}$

b) $xy+yz$

c) $2x-y-z$

d) $p^2-q$

2. Simplify:

a) $(a+2b)^3-3(a+2b)^2(a-2b)+3(a+2b)(a-2b)^2-(a-2b)^2$

b) $(5x-8)^3-(3x-8)^3-6x(5x-8)(3x-8)$

c) $(a-b+c)^3+(a+b-c)^3+6a[a^2-(b-c)^2]$

d) $(5x-2)^3+(3-4x)^3+3(x+1)(5x-2)(3-4x)$

3. Find the value of $a^3+b^3$ when $a+b=6$ and $ab=7$

4. Find the value of $a^3-b^3$ when $a-b=4$ and $ab=2$

5. If $a+\dfrac{1}{a}=3$ , show that, $a^3+(\dfrac{1}{a})^3=18$

6. If $\dfrac{a^2+1}{a}=4$ show that $\dfrac{a^6+1}{a^3}=52$

7. If $x-\dfrac{1}{x}=p$ show that $x^3-\dfrac{1}{x^3}=p^3+3p$

8. If $\dfrac{a^2-1}{a}=1$ find the value of $\dfrac{a^6+1}{a^3}$

Formula:

$(a+b)(a^2-ab+b^2)=a^3+b^3$

Proof:

$(a+b)(a^2-ab+b^2)=a(a^2-ab+b^2)+b(a^2-ab+b^2) \\ =a^3-a^2b+ab^2+a^2b-ab^2+b^3 \\ ==a^3+b^3$

Conversely, $a^3+b^3=(a+b)(a^2-ab+b^2)$ . Hence, any expression of the form $a^3+b^3$ can be resolved into factors.

Example: Multiply $x^4-x^2+1$ by $x^2+1$

Solution:

Putting $a$ for $x^2$ and $b$ for $1$ we have,

$x^4-x^2+1=(x^2)^2-x^2.1+1^2=a^2-ab+b^2$

Hence, $(x^2+1)(x^4-x^2+1)=(a+b)(a^2-ab+b^2) \\ =a^3+b^3 \\ =(x^2)^3+1^3=x^6+1$

Example: Resolve into factors $a^3b^3+8c^3$

Solution:

$a^3b^3+8c^3=(ab)^3+(2c)^3 \\ =(ab+2c)[(ab)^2-(ab)(2c)+(2c)^2] \\ =(ab+2c)(a^2b^2-2abc+4c^2)$

Example: Resolve into factors $64p^3+125$

Solution:

$64p^3+125=(4p)^3+(5)^3 \\ =(4p+5)[(4p)^2-(4p)(5)+(5)^2] \\ =(4p+5)(16p^2-20p+25)$

Formula:

$(a-b)(a^2+ab+b^2)=a^3-b^3$

Proof:

$(a-b)(a^2+ab+b^2)=a(a^2+ab+b^2)-b(a^2+ab+b^2) \\ =a^3+a^2b+ab^2-a^2b-ab^2-b^3 \\ =a^3-b^3$

Conversely, $a^3-b^3=(a-b)(a^2+ab+b^2)$ . Hence, any expression of the form $a^3-b^3$ can be resolved into factors.

Example: Multiply $16a^2+20ab+25b^2$ by $4a-5b$

Solution:

Putting $x$ for $4a$ and $y$ for $5b$ we have,

$16a^2+20ab+25b^2=(4a)^2+(4a)(5b)+(5b)^2 \\ =x^2+xy+y^2$

$\therefore$ $(4a-5b)( 16a^2+20ab+25b^2) \\ =(x-y)(x^2+xy+y^2) \\ =x^3-y^3 \\ =(4a)^3-(5b)^3 \\ =64a^3-125b^3$

Example: Resolve into factors $64x^3-a^3y^6$

Solution:

$64x^3-a^3y^6=(4x^2)^3-(ay^2)^3 \\ =(4x^2-1)[(4x^2)^2+(4x^2)(ay^2)+(ay^2)^2] \\ =(4x^2-ay^2)(16x^4+4ax^2y^2+a^2y^4)$

Example: Resolve into factors $343x^3-8y^6$

Solution:

$343x^3-8y^6=(7x)^3-(2y^2)^3 \\ =(7x-2y^2)[(7x)^2+(7x)(2y^2)+(2y^2)^2] \\ =(7x-2y^2)(49x^2+14xy^2+4y^4)$

Exercise 2:

1. Multiply:

a) $x^2-x+1$ by $x+1$

b) $1-2x+4x^2$ by $1+2x$

c) $1+2x+4x^2$ by $1-2x$

d) $x^2+3x+9$ by $x-3$

2. Resolve into factors:

a) $1331a^3b^6x^9+729c^3y^6z^9$

b) $27a^3+343y^3$

c) $125a^2 -1$

d) $1-512k^3$

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