CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Factorize : $$a^{6}-7a^{3}-8$$


A
(a2)(a+1)(a2+2a+4)(a2a+1)
loader
B
(ab)(a+1)(a2+a+4)(a2a1)
loader
C
(a)(a+1)(a2+2ab+4)(a2a+1)
loader
D
(a2)(a+1)(a2+a+b)(a2b+1)
loader

Solution

The correct option is A $$\left ( a-2 \right )\left ( a+1 \right )\left ( a^{2}+2a+4 \right )\left ( a^{2}-a+1 \right )$$
$$a^{6}-7a^{3}-8$$
$$=a^6-8a^3+a^3-8$$
$$=a^3(a^3-8)+1(a^3-8)$$
$$=(a^3-8)(a^3+1)$$
$$=(a^3-2^3)(a^3+1^3)$$
Using,$$(a^3-b^3)=(a-b)(a^2+ab+b^2)$$
and    $$(a^3+b^3)=(a+b)(a^2-ab+b^2)$$
$$=\left ( a-2 \right )\left ( a+1 \right )\left ( a^{2}+2a+4 \right )\left ( a^{2}-a+1 \right) $$

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More



footer-image