Cube of a Number
- A number whose power is $3$.
- The cube of $x$ is $(x\times x\times x)$ denoted as $x^3$.
$ii)$ $(-2)^3$ $=(-2)\times (-2)\times (-2)$ $=-8$
$iv)$ $\left( \frac{-5}{3} \right)^3$ $=\left( \frac{-5}{3} \right)$ $\left( \frac{-5}{3} \right)$ $\left( \frac{-5}{3} \right)$ $=\frac{-125}{27}$
- A number's cube is the result of multiplying it by three.
Examples:
$iii)$ $\left( \frac{5}{3} \right)^3$ $=\left( \frac{5}{3} \right)\left( \frac{5}{3} \right)\left( \frac{5}{3} \right)$ $=\frac{125}{27}$
Perfect Cube of a Number
- A natural number which is a cube of another number is called a perfect cube of that number.
$1,8,27,64,125,216,347,512,749$, and $1000$ are perfect cubes of $1,2,3,4,5,6,7,8,9$, and $10$ respectively.
Cube Root of a Positive Real Number
- The cube root of a positive real number is a number that when multiplied three times gives the same number.
- It is denoted by $\sqrt[3]{}$ or $\left( \text{ } \right)^{1/3}$
Example:
$i)$ The cube root of $x$ is $\sqrt[3]{x}$ or $\left( x \right)^{1/3}$
$ii)$ The cube root of $27$ is $\sqrt[3]{27}$ $=\sqrt[3]{3\times 3\times 3}=3$
$iii)$ The cube root of $64$ is $4$.
$\sqrt[3]{64}=\sqrt[3]{4\times 4\times 4}$ $=\left( 4^3 \right)^{1/3}=4$
The principal root of a positive real number is $nth$.
For any two positive real numbers $x, y$ and a natural number $n$, if $y^n=x$ then $y$ is called the principal $n^{th}$ root of $x$ and is represented by
$y=$ $\sqrt[n]{x}$
Where "$\sqrt {\text{ }}$" is the symbol of the principal $n^{th}$ root and the principal $n^{th}$ root of a positive real number is the positive root.
In this case, $x$ is called the radicand and $n$ is called the index of the root.
It is also called the perfect $n^{th}$ root of a positive real number.
Remember that:
$n$ is a positive integer and we have defined only the principal $n^{th}$ root of a positive real number.$\sqrt[n]{x}$ is also written as $\left( x \right)^{1/n}$
For example:
$\sqrt 4$ $=2$, $\sqrt[3]{27}$ $=3$, $\sqrt[4]{256}$ $=4$, $\sqrt[5]{3125}$ $=5$, and $\sqrt[6]{46656}$ $=6$ are the principal square root, cube root, $4^{th}$ root, $5^{th}$ root, and 6^{th} root of $4$, $27$, $256$, $3125$, and $46656$ respectively.
Properties of $n^{th}$ root of a positive real number
$1)$ $\sqrt[n]{x}$ $=y\implies x$ $=y^n$$2)$ $\left( \sqrt[n]{x} \right)^n=x$
$3)$ $\sqrt[n]{xy}$ $=\sqrt[n]{x}\text{ }\sqrt[n]y$
$4)$ $\frac{\sqrt[n] x}{\sqrt[n] x}$ $=1$, $(x\neq 0)$
Simplify
$i)$ $\sqrt[4]{256a^4b^4}$Solution:
$\left( 256a^4b^4 \right)^{1/4}$
$=(4\times 4\times 4\times 4\times a^4\times b^4)^{1/4}$
$\left[ \left( 4ab \right)^4 \right]^{1/4}$
$=4ab$
$ii)$ $\sqrt[3]{\frac{64\text{ }a^3\text{ }b^9}{216\text{ }c^6\text{ }d^{18}}}$
Solution:
$\sqrt[3]{\frac{64\text{ }a^3\text{ }b^9}{216\text{ }c^6\text{ }d^{18}}}$
$=\left( \frac{4^3\text{ }a^3\text{ }b^9}{6^3\text{ }c^6\text{ } d^{18}} \right)^{1/3}$
$=\left[ \left( \frac{4\text{ }a\text{ }b^2}{6\text{ }c^2\text{ }d^6} \right)^3 \right]^{1/3}$
$=\frac{2ab^2}{3c^2d^6}$
Rational Exponent
For any positive real number $x$ and natural numbers $m$, $n$ $(m,n\gt 1)$, $x^{m/n}$, is defined as$x^{m/n}$ $=\sqrt[n]{x^{m}}$
In other words, $x^{m/n}$ means the $n^{th}$ root of $x^m$.
For zero and negative rational powers
we define as
$x^0$ $=1$
$x^{-m/n}$ $=\sqrt[n] {x^{-m}}$
If $x,y$ are any two positive real numbers $\frac{m}{n}$, $\frac{k}{l}$ are rational numbers.
then,
$i)$ $x^{m/n}.x^{k/l}$ $=x^{\frac{m}{n}$ $+\frac{k}{l}}$
$ii)$ $\left( x^{\frac{m}{n}} \right)^{\frac{k}{l}}$ $=x^{\frac{mk}{nl}}$
$iii)$ $\frac{x^{\frac mn}}{x^{\frac kl}}$ $=x^{\frac{m}{n}-\frac{k}{l}}$
$iv)$ $\left( xy \right)^{\frac{m}{n}}$ $=x^\frac{m}{n}.y^\frac{m}{n}$
$v)$ $\left( \frac{x}{y} \right)^\frac{m}{n}$ $=\frac{x^{\frac{m}{n}}}{y^{\frac{m}{n}}}$
$a)$ $\frac{(27)^{\frac{2n}{3}}\times(8)^{-\frac{n}{3}}}{(18)^{-\frac{n}{2}}}$
solution:
$\frac{(3^3)^{\frac{2n}{3}}\times(2^3)^{-\frac{n}{3}}}{(2\times 3^2)^{-\frac{n}{2}}}$ $=\frac{3^{2n}\times 2^{-n}}{2^{-\frac{n}{2}}\times 3^{-n}}$
$3^{2n}\times3^n\times 2^{-n}\times 2^{+n/2}$
$3^{3n}\times 2^{-1/n}$ $=\frac{3^{3n}}{2^n}$
$b)$ $\left( \frac{a^p}{a^q} \right)^{-p-q}$ $\times$ $\left( \frac{a^q}{a^r} \right)^{-r-q}$ $\times \left( \frac{a^r}{a^p} \right)^{-p-r}$
Solution:
$\left( \frac{a^q}{a^p} \right)^{p+q}$ $\left( \frac{a^r}{a^q} \right)^{r+q}$ $\left( \frac{a^p}{a^r} \right)^{p+r}$
$\left( {a^{q-p}} \right)^{p+q}$ $\left({a^{r-q}} \right)^{r+q}$ $\left({a^{p-r}}\right)^{p+r}$
${a^{(q-p)({p+q})}}$ $\times {a^{(r-q)({r+q})}}$ $\times{a^{(p-r)({p+r})}}$
$a^{q^2-p^2}$ $\times a^{r^2-q^2}$ $\times a^{p^2-r^2}$
$a^{q^2-p^2+r^2-q^2+p^2-r^2}$
$a^0$ $=1$
$iii)$ $4^{3^{2}}$ $\text{ ÷ }$ $4^{2^3}$
Solution:
$4^{3\times3}$ $\text{ ÷ }$ $4^{2\times2\times2}$
$4^9$ $\text{ ÷ }$ $4^8$ $=4^{9-1}$ $=4^1=4$
Surds
- A surd is an irrational radical with a rational radicand.
$1)$ If $\sqrt[n]{a}$ is an irrational number and $a$ is not a perfect $n^{th}$ power then it is called a surd of $n^{th}$ power.
Examples:
$\sqrt {3}$ is a surd of second order
$\sqrt[3]{5}$ is a surd of third order
$\sqrt[4]{6}$ is a surd of fourth order
$\sqrt[5]{10}$ is a surd of the fifth order
Binomial Surd
- An expression of two terms in which at least one term is a surd of second order is called a "binomial surd".
$2+\sqrt {3}$, $\sqrt {5}$ $-3$, $\sqrt {3}$ $+\sqrt {2}$ are binomial surds
Rationalizing Factor
- When the product of two irrational expressions is rational, then each is said to be a rationalizing factor of the other or conjugate of the other
- Expression of the type $a\sqrt{x}-b\sqrt{y}$. and $a\sqrt{x}+b\sqrt{y}$, $x$ and $y$ are not perfect squares, $a\neq 0$, $b\neq 0$ taken together, are binomial surds and each is called the conjugate of the other.
- The product of a conjugate pair of binomial surds is a rational number.
Write $\frac{1}{ax-b\sqrt y}$ is a form that does not contain a term with a radical sign in the denominator.
$\frac{1}{ax-b\sqrt y}$ $\times$ $\frac{ax+b\sqrt y}{ax-b\sqrt y}$
$=\frac {ax+b\sqrt y}{(ax)^2-(b\sqrt y)^2}$
$=\frac {ax+b\sqrt y}{a^2x^2-b^2y}$
Rationalization
The process of changing $\frac{1}{ax-b\sqrt y}$ into an expression that does not have a term with a radical sign in the denominator is called "rationalization".
$1.$ If $\frac{1}{b}$ $=2+\sqrt{3}$, find the value of $b^4$ $+\frac{1}{b^4}$
Solution:
$\frac{1}{b}$ $=2$ $+\sqrt{3}$
Squaring on both sides
$\left( \frac{1}{b} \right)^2$ $=\left( 2+\sqrt{3} \right)^2$
$\frac{1}{b^2}$ $=4+4\sqrt{3}+3$
$\frac{1}{b^2}$ $=7+4\sqrt{3}$
Again squaring on both sides
$\left( \frac{1}{b^2} \right)^2$ $=\left( 7+4\sqrt{3} \right)^2$
$\frac{1}{b^4}$ $=49+56\sqrt{3}+48$
$\frac{1}{b^4}$ $=97+56$ $\sqrt{3}$ $(i)$
Given equation: $\frac{1}{b}$ $=2+\sqrt{3}$
$\frac{1}{b}$ $=\left( 2+\sqrt{3} \right)$ $\times $ $\left( \frac{2-\sqrt{3}}{2-\sqrt{3}} \right)$ $=\frac{2-\sqrt{3}}{4-3}$ $=2-\sqrt{3}$
$b=2-\sqrt{3}$
$b^2$ $=4-4$ $\sqrt{3}+3$
$b^2$ $=7-4$ $\sqrt{3}$
$b^4$ $=49-56$ $\sqrt{3}+48$
$b^4$ $=97-56$ $\sqrt{3}$ $(ii)$
add $(i)$ and $(ii)$
$b^4+\frac{1}{b^4}$ $=97-56$ $\sqrt{3}$ $+97+56$ $\sqrt{3}$
$b^4+$ $\frac{1}{b^4}$ $=194$ Ans.
$2.$ If $x=$ $\sqrt{2}$ $+\sqrt{3}$, find the value of $x^4$ $+\frac{1}{x^4}$
Solution:
$x=$ $\sqrt{2}$ $+\sqrt{3}$
Squaring on both sides
$x^2$ $=2+2$ $\sqrt{6}$ $+3$
$x^2$ $=5+2$ $\sqrt{6}$
Again squaring on both sides
$x^4$ $=25+20$ $\sqrt{6}$ $+24$
$x^4$ $=49+20$ $\sqrt{6}$ $(i)$
$x=$ $\sqrt{2}$ $+\sqrt{3}$
$\frac{1}{x}$ $=\frac{1}{\sqrt{2}+\sqrt{3}}$
$\frac{1}{x}$ $=\frac{1}{\sqrt{2}+\sqrt{3}}$ $\times$ $\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}$
$\frac{1}{x}$ $=\frac{\sqrt{2}-\sqrt{3}}{2-3}$
$\frac{1}{x}$ $=\frac{\sqrt{2}-\sqrt{3}}{-1}$
$\frac{1}{x}$ $=\sqrt{3}-\sqrt{2}$
$\frac{1}{x}$ $=\sqrt{3}$ $-\sqrt{2}$
$\frac{1}{x^2}$ $=3-2$ $\sqrt{6}+2$
$\frac{1}{x^2}$ $=5-2$ $\sqrt{6}$
$\frac{1}{x^4}$ $=25-20$ $\sqrt{6}+24$
$\frac{1}{x^4}$ $=49-20$ $\sqrt{6}$ $(ii)$
add $(i)$ and $(ii)$
$\frac{1}{x^4}$ $=49+20$ $\sqrt{6}$ $+49-20$ $\sqrt{6}$
$x^4$ $+\frac{1}{x^4}$ $=90$ Ans.
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