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Login Now**Solution I:**

x^{2} + y^{2} is even

We know that,

(x+y)^{2} = x^{2} + 2xy + y^{2}

(x+y)^{2} = x^{2} + y^{2} + 2xy

Since, 2xy is always even and we have given (x^{2} + y^{2}) is even. Addition of both even number results even number. So,

(x+y)^{2} is even. Eventually, x + y is also even.

**Solution II:**

Let P(n) = 1^{3} + 2^{3} + 3^{3} + ………………. + n^{3}

Basic Step:

P(1) = \(\frac{1^2 (1 + 1)^2}{4}\) = 1

So, P(1) is true

Induction HypothesisL

Suppose P(k) is true.

P(k) = 1^{3} + 2^{3} + 3^{3} + ………………. + k^{3} = \(\frac{k^2 (k + 1)^2 }{4}\)

Inductive Step:

Now, we wish to show P(k+1) is true for this, we add (k+1)^{3} on both sides then

1^{3} + 2^{3} + 3^{3} + ………………. + k^{3} + (k+1)^{3} = \(\frac{k^2 (k + 1)^2 }{4} + (k+1)^3\)

= \(\frac{k^2 (k + 1)^2 }{4} + \frac{4(k+1)^3}{4}\)

Taking (k+1)^{2} common from both terms,

= \((k + 1)^2 \left \{ \frac{k^2}{4} + \frac{4(k+1)}{4} \right \}\)

= \((k + 1)^2 \left \{ \frac{k^2 }{4} + \frac{4k+4)}{4} \right \}\)

= \((k + 1)^2 \left \{ \frac{k^2 + 4k + 4}{4} \right \}\)

= \(\frac{(k + 1)^2 (k + 2)^2}{4}\)

= \(\frac{(k + 1)^2 (k + 1 + 1)^2}{4}\)

Therefore, P(k+1) = \(\frac{(k + 1)^2 (k + 1 + 1)^2}{4}\) is true.

Thus, by mathematical induction, P(n) is true for all n.

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