8th Class Mathematics Cubes and Cube Roots Cube and Cube Roots

CUBE & CUBE ROOTS

FUNDAMENTALS

Cube and cube root

• Cube:- If y is a non-zero number, then \[y\times y\times y\] written as y³ is called the cube of y or simply y cubed.

e.g.,  

(i) \[{{\left( 5 \right)}^{3}}=5\times 5\times 5=125.\] Thus, Cube of 5 is 125.

(ii) \[{{\left( 9 \right)}^{3}}=9\times 9\times 9=729\]. Thus,

• Perfect cube:- A natural number n is a perfect cube if it is the cube of some natural number.

Or

Natural number n is a perfect cube if there exists a natural number whose cube is n

i.e. \[n={{x}^{3}}\]

e.g.,(i) 343 is a perfect cube, because there is a natural number 7 such that

\[343=7\times 7\times 7={{7}^{3}}\]

e.g., (ii) \[{{4}^{3}}=4\times 4\times 4=64\]

\[{{5}^{3}}=5\times 5\times 5=125\]

\[{{9}^{3}}=9\times 9\times 9\times =723\]

Properties of perfect cube:

• If 'n' is even, then \[{{n}^{3}}\] is also even.
• If 'n' is odd, then \[{{n}^{3}}\] is also odd.
• If 'm' is even and 'n^’ is odd, then \[{{m}^{3}}\times {{n}^{3}}\]is even.
• If a number's units place has digit 1, 4, 5, 6, then its Cube also ends in the same digit
• Cube of negative number is negative

\[{{(-1)}^{3}}=-1,{{(-~9)}^{3}}=-729\]

Some Shortcuts to find cubes

• Column method:- Let x = ab (where a is tens digit and b is units digit)

Be a 2 digit natural number.

Then \[{{(a+b)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}\]

e.g.. Find the cube of 26 by using column method.

Solution:- By Using column method , we have

\[\therefore {{\left( 26 \right)}^{3}}=17576\]

Cube root:- If 'x^’ is a perfect cube and for some integers y, \[x={{y}^{3}}\], then the number 'y' is called cube root of 'x\ It is denoted by \[y=\sqrt[3]{x}or\,{{x}^{\frac{1}{3}}}.\]

Example:-

\[27={{3}^{3}}\]                     \[\therefore \sqrt[3]{27}=3\]

\[729={{9}^{3}}\]                   \[\therefore \sqrt[3]{729}=9\]

\[-1000={{\left( -10 \right)}^{3}}\]         \[\therefore \sqrt[3]{1000}=10\]

\[0.008={{\left( 0.2 \right)}^{3}}\]                     \[\therefore \sqrt[3]{0.008}=0.2\]

\[\frac{1}{125}={{\left( \frac{1}{5} \right)}^{3}}\]                     \[\therefore \sqrt[3]{\frac{1}{125}}=\frac{1}{5}\]

Method to find the cube root of a number: 

• Prime factorization method:-

Follow these steps:

• Resolve the given number into its prime factors
• Make triplets of equal factors.
• Take the product of the prime factors, choosing one factor out of every triplet.

e.g.,(i) Find the cube root of 2744

solution:- By prime factorization we get

\[2744=2\times 2\times 2\times 7\times 7\times 7\]

\[\therefore \sqrt[3]{2744}=2\times 7=14\]

e.g.,(ii) What is the smallest number by which 3087 must be divided so that the quotient is a perfect cube?

Solution:- Resolving 3087 in to prime factors, we get \[3087=3\times 3\times 7\times 7\times 7.\] By grouping the factors clearly, if we divide 3087 by \[3\times 3=9\] the quotient would be \[7\times 7\times 7\] which is a perfect cube.

Note:-

• Cube root of a negative number is negative, i.e., \[\sqrt[3]{-{{x}^{3}}}=-x\]
• Cube root of product of two integers, is product of their cube roots: \[\sqrt[3]{x.y}=\sqrt[3]{x.}\sqrt[3]{y}\]
• Cube root of the a rational number, is cube root of Numerator divided by cube root of Denominator: \[\sqrt[3]{\frac{x}{y}}=\frac{\sqrt[3]{x.}}{\sqrt[3]{y}}(y\ne 0)\]

Remember these identities

• \[\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}\]
• \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]
• \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]
• \[{{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}\]
• \[{{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}\]
• \[{{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}})\]
• \[{{a}^{3}}+{{b}^{3}}=(a+b)({{a}^{2}}-ab+{{b}^{2}})\]
• \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\] \[=(a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca)\]