Question

Factorise: 14x^6 – 45x^3y^3 – 14y^6​

Answer 1

Answer:

$(7x^{3} +2y^{3} )(2x^{3} -7y^{3} )$

Step-by-step explanation:

$14x^{6} -45x^{3} y^{3} -14y^{6}$

$=14x^{6} -49x^{3} y^{3}+4x^{3} y^{3} -14y^{6}$

$=7x^{3} (2x^{3} -7y^{3} )+2y^{3}(2x^{3}  -7y^{3} )$

$=(7x^{3} +2y^{3} )(2x^{3} -7y^{3} )$

Answer 2

it is given that , 14x^6 - 45x^3y^3 - 14y^6

= 14x^6 - 49x^3y^3 + 4x^3y^3 - 14y^6

= 7x^3(2x^3 - 7y^3) + 2y^3(2x^3 - 7y^3)

= (7x^3 + 2y^3)(2x^3 - 7y^3)

[ it can also be a solution]

now, for (7x³ + 2y³) , we can write it $(\sqrt[3]{7}x)^3+(\sqrt[3]{2}y)^3$

from algebraic identity,

a³ + b³ = (a + b)(a² - ab + b³)

so, $(\sqrt[3]{7}x)^3+(\sqrt[3]{2}y)^3$ = $(\sqrt[3]{7}x+\sqrt[3]{2}y)(\sqrt[3]{49}x^2+\sqrt[3]{4}y^2-\sqrt[3]{14}xy)$

similarly from algebraic identity,

a³ - b³ = (a - b)(a² + ab + b²)

or, $(2x^3-7y^3)=(\sqrt[3]{2}x)^3-(\sqrt[3]{7}y)^3$

or, $=(\sqrt[3]{2}x-\sqrt[3]{7}y)(\sqrt[3]{4}x^2+\sqrt[3]{49}y^3+\sqrt[3]{14}xy)$

hence, factorisation of 14x^6 - 45x^3y^3 - 14y^6

$=(\sqrt[3]{7}x+\sqrt[3]{2}y)(\sqrt[3]{49}x^2+\sqrt[3]{4}y^2-\sqrt[3]{14}xy)(\sqrt[3]{2}x-\sqrt[3]{7}y)(\sqrt[3]{4}x^2+\sqrt[3]{49}y^3+\sqrt[3]{14}xy)$