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3 years ago

Without actually calculating the cubes, find the value of (16)^3+(-7)^3+(-9)^3

Mathematics

2 answers:

svlad2 [7]3 years ago

7 0

Answer:

(16)^3 = (7 + 9)^3 = 7^3 + 9^3 + 3 x 7^2 x 9 + 3 x 9^2 x 7

=>(16)^3 + (-7)^3 + (-9)^3 = (-7)^3 + (-9)^3 + 7^3 + 9^3 + 3 x 7^2 x 9 + 3 x 9^2 x 7

= [(-7)^3 + 7^3] + [9^3 + (-9)^3] + 3 x 7^2 x 9 + 3x9^2x7

= 0 + 0 + 3 x 7^2 x 9 + 3 x 9^2 x 7

= 3 x 7^2 x 9 + 3 x 9^2 x 7

= 3 x 9 x 7 x (7 + 9)

= 3 x 7 x 9 x 16

= 3 x 7 x 9 x 8 x 2

= 6 x 7 x 8 x 9

= 3024

Hope this helps!

BartSMP [9]3 years ago

6 0

The value i got was 3024

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If sinA=√3-1/2√2,then prove that cos2A=√3/2 prove that

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Answer:

$\boxed{\sf cos2A =\dfrac{\sqrt3}{2}}$

Step-by-step explanation:

Here we are given that the value of sinA is √3-1/2√2 , and we need to prove that the value of cos2A is √3/2 .

Given :-

• $\sf\implies sinA =\dfrac{\sqrt3-1}{2\sqrt2}$

To Prove :-

• $\sf\implies cos2A =\dfrac{\sqrt3}{2}$

Proof :-

We know that ,

$\sf\implies cos2A = 1 - 2sin^2A$

Therefore , here substituting the value of sinA , we have ,

$\sf\implies cos2A = 1 - 2\bigg( \dfrac{\sqrt3-1}{2\sqrt2}\bigg)^2$

Simplify the whole square ,

$\sf\implies cos2A = 1 -2\times \dfrac{ 3 +1-2\sqrt3}{8}$

Add the numbers in numerator ,

$\sf\implies cos2A = 1-2\times \dfrac{4-2\sqrt3}{8}$

Multiply it by 2 ,

$\sf\implies cos2A = 1 - \dfrac{ 4-2\sqrt3}{4}$

Take out 2 common from the numerator ,

$\sf\implies cos2A = 1-\dfrac{2(2-\sqrt3)}{4}$

Simplify ,

$\sf\implies cos2A = 1 -\dfrac{ 2-\sqrt3}{2}$

Subtract the numbers ,

$\sf\implies cos2A = \dfrac{ 2-2+\sqrt3}{2}$

Simplify,

$\sf\implies \boxed{\pink{\sf cos2A =\dfrac{\sqrt3}{2}} }$

Hence Proved !

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Answer:

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Step-by-step explanation:

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2 years ago

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