Find the product for $x = 1, y = 1$
$(2 x + 3 y) (4 x^{2} - 6 x y + 9 y^{2})$

30

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35

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40

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45

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Solution

The correct option is D $35$
We know that one of the formula is $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ .

In the given product $(2 x + 3 y) (4 x^{2} - 6 x y + 9 y^{2})$ , we observe that $a = 2 x$ and $b = 3 y$ , therefore, comparing with the above formula, we get:

$(2 x + 3 y) (4 x^{2} - 6 x y + 9 y^{2}) = (2 x)^{3} + (3 y)^{3} = 8 x^{3} + 27 y^{3}$

Let us substitute $x = 1$ and $y = 1$ in $8 x^{3} + 27 y^{3}$ as shown below:

$(8 \times 1^{3}) + (27 \times 1^{3}) = (8 \times 1) + (27 \times 1) = 8 + 27 = 35$

Hence, $(2 x + 3 y) (4 x^{2} - 6 x y + 9 y^{2}) = 35$ if $x = 1$ and $y = 1$ .

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