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Factorisation Using Cubic Identities

Algebraic Identities - Concepts

Algebraic Identity - Square of Sum of Binomial

Algebraic Identity - Square of difference of Binomial

Algebraic Identity - Dfference of Square

Algebraic Identity -3

Algebraic Identity - Square of Trinomial

Algebraic Identity - Difference of Cubes

Algebraic Identity - Sum of Cubes

Algebraic Identity -Cube of Sum of Binomial

Algebraic Identity -Cube of Difference of Binomial

Algebraic Identity- Conditional

Factorisation Using Square Identities

Factorisation Using Cubic Identities

Class - Clerk-cum-Cashier Subjects

Concept Explanation

Factorisation Using Cubic Identities

Factorization of Algebraic Expressions Expressible as a Perfect Cube:

In order to factorize algebraic expressions expressible as a perfect cube, we use the following expressions.

$large (i);a^{3}+3a^2b+3ab^2+b^{3}=(a+b)^{3}=(a+b)(a+b)(a+b)$

$large (ii);a^{3}-3a^2b+3ab^2+b^{3}=(a-b)^{3}=(a-b)(a-b)(a-b)$

Illustration: Factorise the following

$large (i);8x^3+36x^2y+54xy^2+27y^3$

$large =(2x)^{3}+3times (2x)^2times 3y+3times 2xtimes (3y)^2+(3y)^{3}$

$large =(2x+3y)^{3}$ $large =(2x+3y)(2x+3y)(2x+3y)$

Factorization of Algebraic Expressions using Conditional Identity:

In order to factorize algebraic expressions expressible as the following expressions.

$large ;a^{3}+b^3+c^3-3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

Illustration: Factorise the following

$large 8x^3+y^3+27z^3-18xyz$

$large =(2x)^{3}+y^3+(3z)^3-3 times (2x)times (y) times (3z)$

$=(2x+y+3z)((2x)^{2}+y^2+(3z)^2-2x times (y)- (y)times (3z)-(3z) times (2x))$

$large =(2x+y+3z)(4x^2+y^2+9z^2-2xy- 3yz -6xz)$

Sample Questions

(More Questions for each concept available in Login)

Question : 1

The factors of $8a^3+b^3-6ab+1: : are$
A. $(2a+b-1)(4a^2+b^2+1-3ab-2a)$
B. $(2a-b+1)(4a^2+b^2-4ab+1-2a+b)$
C. $(2a+b+1)(4a^2+b^2+1-2ab-b-2a)$
D. $(2a-1+b)(4a^2+1-4a-b-2ab)$
Right Option : C
View Explanation

Question : 2

One of the factor of $a^{3}+8b^{2}-64c^{3}+24abc ;;is:$
A. a + 2b - 4c
B. a - 2b + 4c
C. a + 2b + 4c
D. a - 2b - 4c
Right Option : A
View Explanation

Question : 3

$a^1^2-1$ can be factorised as :
A. $large =(a-1)(a^2+1-a)(a-1)(a^2+a+1)(a^2+1)(a^4+1-a^2)$
B. $large =(a +1)(a^2+1+a)(a-1)(a^2+a+1)(a^2+1)(a^4+1-a^2)$
C. $large =(a +1)(a^2+1+a)(a+1)(a^2+a+1)(a^2+1)(a^4+1-a^2)$
D. $large =(a +1)(a^2+1-a)(a-1)(a^2+a+1)(a^2+1)(a^4+1-a^2)$
Right Option : D
View Explanation

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Chapters

Partnership and Mixture

Problems On Boats And Stream

Rules For Divisibility

Number System

Cube and Cube Roots

Number Theory

Square and Square Root

Permutation and Combination

Algebraic Identities

Time, Speed and Distance

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Andhra Bank Clerk-cum-Cashier Test Series

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Attention !!!!!

Factorisation Using Cubic Identities

Factorisation Using Cubic Identities

Factorization of Algebraic Expressions Expressible as a Perfect Cube:

Factorization of Algebraic Expressions using Conditional Identity:

Students / Parents Reviews [10]

Eshan Arora

Archit Segal

Om Umang

Hiya Gupta

Ayan Ghosh

Shreya Shrivastava

Manish Kumar

Yatharthi Sharma

Shivam Rana

Barkha Arora