Question

factorize the following :

${64x}^{3} + {27y}^{3} + {8z}^{3} - 72xyz$
​

Answer 1

Answer:

To Factorize 64x³+27y³+8z³-72xyz

Solution:-

$64 {x}^{3} + 27 {y}^{3} + 8 {z}^{3} - 72xyz \\ =(4x)^{3} + ({3y})^{3} + ({2z})^{3} - 72xyz$

we know

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

here

a=4x

b=3y

c=2z

therefore we've to substitute those values in the equation.

•

$(4x+3y+2z) \\ (({4x})²+({3y})²+({2z})²-(4x \times 3y ) \\ - (3y \times 2z) - (2z \times 4x) \\ = (4x+3y+2z)(16 {x}^{2} + 9 {y}^{2} + 4 {z}^{2} ) \\ (- 12xy - 6yz - 8zx)$

therefore the answer is.

• (4x+3y+2z)(16x²+9y²+4z²-12xy-6yz-8zx)

hope it helps