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Which of the following is NOT true if $p^2 = p + 1$ and p is a real number
A. $p^3 = p^2 + p$
B. $p^4 = p^3 + p + 1$
C. $p^3 = 2p + 1$
D. $ p^3 + p^2 = p + 1$
E. $p = \frac{1 } {pāˆ’1}$

I tried factoring it but it did not work.

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    Hint: just manipulate each expression in turn. For example, $p^3=p^2+p$ follows from $p^2=p+1$ by multiplying through by $p$. And so on. ā€“ 2017-01-23

2 Answers 2

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In D you have $p^3+p^2=p+1$, but $p^2=p+1$ so $$p^3+p^2=p^2$$ $$p^3=0$$ $$p=0$$ but this is not true from $p^2=p+1$

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Just manipulate each expression but some cases do follow each other.

As $p^2=p+1$, we have $$p^3=p (p+1) =p^2+p =(p+1) +p =2p+1$$ So $A $ and $C $ are correct.

We have $$p^2-p = p (p-1)=1\Rightarrow p=\frac {1}{p-1} $$, so $E $ follows.

Squaring we get, $$p^4=p^2+2p+1 =(p^2+p)+(p+1) =p^3+p+1$$, so $C $ is correct (from $A $, which is correct, we use $p^3=p^2+p $).

From $C $, we can see clearly that $D $ doesn't follow and is hence the only incorrect option. Hope it helps.

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    i do not get what u did here: Squaring we get, p4=p2+2p+1=(p2+p)+(p+1)=p3+p+1p4=p2+2p+1=(p2+p)+(p+1)=p3+p+1 I know u swuared (p^2), which is p+1, giving us p^2+2p+1, and p^2=(p+1), however where did the (p^2+p) come from???? ā€“ 2017-01-23