Evaluate: $(t - 8) (t + 8)$

t^{2} - 64

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t^{2} - 8

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t + 64

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t^{2} - 4

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Solution

The correct option is B $t^{2} - 64$
By using identity $(a + b) (a - b) = a^{2} - b^{2}$
$(t - 8) (t + 8) = t . t + t .8 - 8. t - 8.8$
= $t^{2} + 8 t - 8 t - 64$
= $t^{2} - 64$

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Similar questions

L t t \to 8 \frac{\sqrt[3]{t} - 2}{t^{2} - 64}

If g = $t^{\frac{2}{3}} + 4 t^{\frac{- 1}{2}}$ , what is the value of g when t = 64?

y^{2} = 4 a x

(a t^{2}, 2 a t)

(a T^{2}, 2 a T)

y^{2} = 4 a x .

(a t^{2}, 2 a t)

(a T^{2}, 2 a T),

y = \frac{(t^{2} - 2)}{(t^{2} + 2)}

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