Question

(iii) (3x +4)^3-(3x-4)^3​

Answer 1

Answer:

$128 + 216 {x}^{2}$

Step-by-step explanation:

${(a + b)}^{3 } = {a}^{3} + {b}^{3} + 3 {a}^{2} b + 3a {b}^{2} \\ {( a - b)}^{3} = {a}^{3} - {b }^{3} - 3 {a}^{2} b + 3 a{b}^{2}$

$a = 3x \\ b = 4$

${( 3x+ 4)}^{3} - {(3x - 4)}^{3} \\ ({(3x)}^{3} + {4}^{3} +3 {(3x)}^{2}4 + 3(3x) {(4)}^{2} ) - ( {(3x)}^{3} - {4}^{3} - 3 {(3x)}^{2} 4 + 3(3x) {(4)}^{2} )$

if you remove the brackets, positive and negative terms will be get cancelled, the remaining terms are as follows

${4}^{3} + {4}^{3} + 3 {(3x)}^{2} 4 + 3 {(3x)}^{2} 4 \\ 64 + 64 + 12(9 {x}^{2} ) + 12(9 {x}^{2} ) \\ 128 + 24(9 {x}^{2} ) \\ 128 + 216 {x}^{2}$

I hope this may be the answer for your question.